Solve by Completing the Square 5x^2+6x-4=0

$5 x^{2} + 6 x - 4 = 0$

Step 1

Add $4$ to both sides of the equation.

$5 x^{2} + 6 x = 4$

Step 2

Divide each term in $5 x^{2} + 6 x = 4$ by $5$ and simplify.

Divide each term in $5 x^{2} + 6 x = 4$ by $5$ .

$\frac{5 x^{2}}{5} + \frac{6 x}{5} = \frac{4}{5}$

Simplify the left side.

Cancel the common factor of $5$ .

Cancel the common factor.

$\frac{5 x^{2}}{5} + \frac{6 x}{5} = \frac{4}{5}$

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

$x^{2} + \frac{6 x}{5} = \frac{4}{5}$

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 $b$ .

${(\frac{b}{2})}^{2} = {(\frac{3}{5})}^{2}$

Step 4

Add the term to each side of the equation.

$x^{2} + \frac{6 x}{5} + {(\frac{3}{5})}^{2} = \frac{4}{5} + {(\frac{3}{5})}^{2}$

Step 5

Simplify the equation.

Simplify the left side.

Simplify each term.

Apply the product rule to $\frac{3}{5}$ .

$x^{2} + \frac{6 x}{5} + \frac{3^{2}}{5^{2}} = \frac{4}{5} + {(\frac{3}{5})}^{2}$

Raise $3$ to the power of $2$ .

$x^{2} + \frac{6 x}{5} + \frac{9}{5^{2}} = \frac{4}{5} + {(\frac{3}{5})}^{2}$

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

$x^{2} + \frac{6 x}{5} + \frac{9}{25} = \frac{4}{5} + {(\frac{3}{5})}^{2}$

Simplify the right side.

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

Simplify each term.

Apply the product rule to $\frac{3}{5}$ .

$x^{2} + \frac{6 x}{5} + \frac{9}{25} = \frac{4}{5} + \frac{3^{2}}{5^{2}}$

Raise $3$ to the power of $2$ .

$x^{2} + \frac{6 x}{5} + \frac{9}{25} = \frac{4}{5} + \frac{9}{5^{2}}$

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

$x^{2} + \frac{6 x}{5} + \frac{9}{25} = \frac{4}{5} + \frac{9}{25}$

To write $\frac{4}{5}$ as a fraction with a common denominator, multiply by $\frac{5}{5}$ .

$x^{2} + \frac{6 x}{5} + \frac{9}{25} = \frac{4}{5} \cdot \frac{5}{5} + \frac{9}{25}$

Write each expression with a common denominator of $25$ , by multiplying each by an appropriate factor of $1$ .

Multiply $\frac{4}{5}$ by $\frac{5}{5}$ .

$x^{2} + \frac{6 x}{5} + \frac{9}{25} = \frac{4 \cdot 5}{5 \cdot 5} + \frac{9}{25}$

Multiply $5$ by $5$ .

$x^{2} + \frac{6 x}{5} + \frac{9}{25} = \frac{4 \cdot 5}{25} + \frac{9}{25}$

Combine the numerators over the common denominator.

$x^{2} + \frac{6 x}{5} + \frac{9}{25} = \frac{4 \cdot 5 + 9}{25}$

Simplify the numerator.

Multiply $4$ by $5$ .

$x^{2} + \frac{6 x}{5} + \frac{9}{25} = \frac{20 + 9}{25}$

Add $20$ and $9$ .

$x^{2} + \frac{6 x}{5} + \frac{9}{25} = \frac{29}{25}$

Step 6

Factor the perfect trinomial square into ${(x + \frac{3}{5})}^{2}$ .

${(x + \frac{3}{5})}^{2} = \frac{29}{25}$

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.

$x + \frac{3}{5} = \pm \sqrt{\frac{29}{25}}$

Simplify $\pm \sqrt{\frac{29}{25}}$ .

Rewrite $\sqrt{\frac{29}{25}}$ as $\frac{\sqrt{29}}{\sqrt{25}}$ .

$x + \frac{3}{5} = \pm \frac{\sqrt{29}}{\sqrt{25}}$

Simplify the denominator.

Rewrite $25$ as $5^{2}$ .

$x + \frac{3}{5} = \pm \frac{\sqrt{29}}{\sqrt{5^{2}}}$

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

$x + \frac{3}{5} = \pm \frac{\sqrt{29}}{5}$

Subtract $\frac{3}{5}$ from both sides of the equation.

$x = \pm \frac{\sqrt{29}}{5} - \frac{3}{5}$

Step 8

The result can be shown in multiple forms.

Exact Form:

$x = \pm \frac{\sqrt{29}}{5} - \frac{3}{5}$

Decimal Form:

$x = 0.47703296 \dots, - 1.67703296 \dots$

Solve by Completing the Square 5x^2+6x-4=0

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