Solve by Completing the Square 2x^2-5x+3=0

$2 x^{2} - 5 x + 3 = 0$

Step 1

Subtract $3$ from both sides of the equation.

$2 x^{2} - 5 x = - 3$

Step 2

Divide each term in $2 x^{2} - 5 x = - 3$ by $2$ and simplify.

Divide each term in $2 x^{2} - 5 x = - 3$ by $2$ .

$\frac{2 x^{2}}{2} + \frac{- 5 x}{2} = \frac{- 3}{2}$

Simplify the left side.

Simplify each term.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 x^{2}}{2} + \frac{- 5 x}{2} = \frac{- 3}{2}$

Divide $x^{2}$ by $1$ .

$x^{2} + \frac{- 5 x}{2} = \frac{- 3}{2}$

Move the negative in front of the fraction.

$x^{2} - \frac{5 x}{2} = \frac{- 3}{2}$

Simplify the right side.

Move the negative in front of the fraction.

$x^{2} - \frac{5 x}{2} = - \frac{3}{2}$

Step 3

To create a trinomial square on the left side of the equation, find a value that is equal to the square of half of $b$ .

${(\frac{b}{2})}^{2} = {(- \frac{5}{4})}^{2}$

Step 4

Add the term to each side of the equation.

$x^{2} - \frac{5 x}{2} + {(- \frac{5}{4})}^{2} = - \frac{3}{2} + {(- \frac{5}{4})}^{2}$

Step 5

Simplify the equation.

Simplify the left side.

Simplify each term.

Use the power rule ${(a b)}^{n} = a^{n} b^{n}$ to distribute the exponent.

Apply the product rule to $- \frac{5}{4}$ .

$x^{2} - \frac{5 x}{2} + {(- 1)}^{2} {(\frac{5}{4})}^{2} = - \frac{3}{2} + {(- \frac{5}{4})}^{2}$

Apply the product rule to $\frac{5}{4}$ .

$x^{2} - \frac{5 x}{2} + {(- 1)}^{2} \frac{5^{2}}{4^{2}} = - \frac{3}{2} + {(- \frac{5}{4})}^{2}$

Raise $- 1$ to the power of $2$ .

$x^{2} - \frac{5 x}{2} + 1 \frac{5^{2}}{4^{2}} = - \frac{3}{2} + {(- \frac{5}{4})}^{2}$

Multiply $\frac{5^{2}}{4^{2}}$ by $1$ .

$x^{2} - \frac{5 x}{2} + \frac{5^{2}}{4^{2}} = - \frac{3}{2} + {(- \frac{5}{4})}^{2}$

Raise $5$ to the power of $2$ .

$x^{2} - \frac{5 x}{2} + \frac{25}{4^{2}} = - \frac{3}{2} + {(- \frac{5}{4})}^{2}$

Raise $4$ to the power of $2$ .

$x^{2} - \frac{5 x}{2} + \frac{25}{16} = - \frac{3}{2} + {(- \frac{5}{4})}^{2}$

Simplify the right side.

Simplify $- \frac{3}{2} + {(- \frac{5}{4})}^{2}$ .

Simplify each term.

Use the power rule ${(a b)}^{n} = a^{n} b^{n}$ to distribute the exponent.

Apply the product rule to $- \frac{5}{4}$ .

$x^{2} - \frac{5 x}{2} + \frac{25}{16} = - \frac{3}{2} + {(- 1)}^{2} {(\frac{5}{4})}^{2}$

Apply the product rule to $\frac{5}{4}$ .

$x^{2} - \frac{5 x}{2} + \frac{25}{16} = - \frac{3}{2} + {(- 1)}^{2} \frac{5^{2}}{4^{2}}$

Raise $- 1$ to the power of $2$ .

$x^{2} - \frac{5 x}{2} + \frac{25}{16} = - \frac{3}{2} + 1 \frac{5^{2}}{4^{2}}$

Multiply $\frac{5^{2}}{4^{2}}$ by $1$ .

$x^{2} - \frac{5 x}{2} + \frac{25}{16} = - \frac{3}{2} + \frac{5^{2}}{4^{2}}$

Raise $5$ to the power of $2$ .

$x^{2} - \frac{5 x}{2} + \frac{25}{16} = - \frac{3}{2} + \frac{25}{4^{2}}$

Raise $4$ to the power of $2$ .

$x^{2} - \frac{5 x}{2} + \frac{25}{16} = - \frac{3}{2} + \frac{25}{16}$

To write $- \frac{3}{2}$ as a fraction with a common denominator, multiply by $\frac{8}{8}$ .

$x^{2} - \frac{5 x}{2} + \frac{25}{16} = - \frac{3}{2} \cdot \frac{8}{8} + \frac{25}{16}$

Write each expression with a common denominator of $16$ , by multiplying each by an appropriate factor of $1$ .

Multiply $\frac{3}{2}$ by $\frac{8}{8}$ .

$x^{2} - \frac{5 x}{2} + \frac{25}{16} = - \frac{3 \cdot 8}{2 \cdot 8} + \frac{25}{16}$

Multiply $2$ by $8$ .

$x^{2} - \frac{5 x}{2} + \frac{25}{16} = - \frac{3 \cdot 8}{16} + \frac{25}{16}$

Combine the numerators over the common denominator.

$x^{2} - \frac{5 x}{2} + \frac{25}{16} = \frac{- 3 \cdot 8 + 25}{16}$

Simplify the numerator.

Multiply $- 3$ by $8$ .

$x^{2} - \frac{5 x}{2} + \frac{25}{16} = \frac{- 24 + 25}{16}$

Add $- 24$ and $25$ .

$x^{2} - \frac{5 x}{2} + \frac{25}{16} = \frac{1}{16}$

Step 6

Factor the perfect trinomial square into ${(x - \frac{5}{4})}^{2}$ .

${(x - \frac{5}{4})}^{2} = \frac{1}{16}$

Step 7

Solve the equation for $x$ .

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

$x - \frac{5}{4} = \pm \sqrt{\frac{1}{16}}$

Simplify $\pm \sqrt{\frac{1}{16}}$ .

Rewrite $\sqrt{\frac{1}{16}}$ as $\frac{\sqrt{1}}{\sqrt{16}}$ .

$x - \frac{5}{4} = \pm \frac{\sqrt{1}}{\sqrt{16}}$

Any root of $1$ is $1$ .

$x - \frac{5}{4} = \pm \frac{1}{\sqrt{16}}$

Simplify the denominator.

Rewrite $16$ as $4^{2}$ .

$x - \frac{5}{4} = \pm \frac{1}{\sqrt{4^{2}}}$

Pull terms out from under the radical, assuming positive real numbers.

$x - \frac{5}{4} = \pm \frac{1}{4}$

The complete solution is the result of both the positive and negative portions of the solution.

First, use the positive value of the $\pm$ to find the first solution.

$x - \frac{5}{4} = \frac{1}{4}$

Move all terms not containing $x$ to the right side of the equation.

Add $\frac{5}{4}$ to both sides of the equation.

$x = \frac{1}{4} + \frac{5}{4}$

Combine the numerators over the common denominator.

$x = \frac{1 + 5}{4}$

Add $1$ and $5$ .

$x = \frac{6}{4}$

Cancel the common factor of $6$ and $4$ .

Factor $2$ out of $6$ .

$x = \frac{2 (3)}{4}$

Cancel the common factors.

Factor $2$ out of $4$ .

$x = \frac{2 \cdot 3}{2 \cdot 2}$

Cancel the common factor.

$x = \frac{2 \cdot 3}{2 \cdot 2}$

Rewrite the expression.

$x = \frac{3}{2}$

Next, use the negative value of the $\pm$ to find the second solution.

$x - \frac{5}{4} = - \frac{1}{4}$

Move all terms not containing $x$ to the right side of the equation.

Add $\frac{5}{4}$ to both sides of the equation.

$x = - \frac{1}{4} + \frac{5}{4}$

Combine the numerators over the common denominator.

$x = \frac{- 1 + 5}{4}$

Add $- 1$ and $5$ .

$x = \frac{4}{4}$

Divide $4$ by $4$ .

$x = 1$

The complete solution is the result of both the positive and negative portions of the solution.

$x = \frac{3}{2}, 1$

Solve by Completing the Square 2x^2-5x+3=0

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