Solve by Completing the Square 5x^2=60-20x

$5 x^{2} = 60 - 20 x$

Step 1

Simplify the equation into a proper form to complete the square.

Reorder $60$ and $- 20 x$ .

$5 x^{2} = - 20 x + 60$

Add $20 x$ to both sides of the equation.

$5 x^{2} + 20 x = 60$

Step 2

Divide each term in $5 x^{2} + 20 x = 60$ by $5$ and simplify.

Divide each term in $5 x^{2} + 20 x = 60$ by $5$ .

$\frac{5 x^{2}}{5} + \frac{20 x}{5} = \frac{60}{5}$

Simplify the left side.

Simplify each term.

Cancel the common factor of $5$ .

Cancel the common factor.

$\frac{5 x^{2}}{5} + \frac{20 x}{5} = \frac{60}{5}$

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

$x^{2} + \frac{20 x}{5} = \frac{60}{5}$

Cancel the common factor of $20$ and $5$ .

Factor $5$ out of $20 x$ .

$x^{2} + \frac{5 (4 x)}{5} = \frac{60}{5}$

Cancel the common factors.

Factor $5$ out of $5$ .

$x^{2} + \frac{5 (4 x)}{5 (1)} = \frac{60}{5}$

Cancel the common factor.

$x^{2} + \frac{5 (4 x)}{5 \cdot 1} = \frac{60}{5}$

Rewrite the expression.

$x^{2} + \frac{4 x}{1} = \frac{60}{5}$

Divide $4 x$ by $1$ .

$x^{2} + 4 x = \frac{60}{5}$

Simplify the right side.

Divide $60$ by $5$ .

$x^{2} + 4 x = 12$

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 $b$ .

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

Step 4

Add the term to each side of the equation.

$x^{2} + 4 x + {(2)}^{2} = 12 + {(2)}^{2}$

Step 5

Simplify the equation.

Simplify the left side.

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

$x^{2} + 4 x + 4 = 12 + {(2)}^{2}$

Simplify the right side.

Simplify $12 + {(2)}^{2}$ .

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

$x^{2} + 4 x + 4 = 12 + 4$

Add $12$ and $4$ .

$x^{2} + 4 x + 4 = 16$

Step 6

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

${(x + 2)}^{2} = 16$

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.

$x + 2 = \pm \sqrt{16}$

Simplify $\pm \sqrt{16}$ .

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

$x + 2 = \pm \sqrt{4^{2}}$

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

$x + 2 = \pm 4$

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.

$x + 2 = 4$

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

Subtract $2$ from both sides of the equation.

$x = 4 - 2$

Subtract $2$ from $4$ .

$x = 2$

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

$x + 2 = - 4$

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

Subtract $2$ from both sides of the equation.

$x = - 4 - 2$

Subtract $2$ from $- 4$ .

$x = - 6$

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

$x = 2, - 6$

Solve by Completing the Square 5x^2=60-20x

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