Solve x^2div4+left(x+6right)^2div4-20=4(2-(x+6))

$${ x }^{ 2 } \div 4+ { \left(x+6 \right) }^{ 2 } \div 4-20=4(2-(x+6))$$

Solve for x

$$x=\sqrt{39}-7\approx -0.755002002$$

$$x=-\left(\sqrt{39}+7\right)\approx -13.244997998$$

Steps Using the Quadratic Formula

Multiply both sides of the equation by 4.

$$x^{2}+\left(x+6\right)^{2}-80=16\left(2-\left(x+6\right)\right)$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+6\right)^{2}.

$$x^{2}+x^{2}+12x+36-80=16\left(2-\left(x+6\right)\right)$$

Combine x^{2} and x^{2} to get 2x^{2}.

$$2x^{2}+12x+36-80=16\left(2-\left(x+6\right)\right)$$

Subtract 80 from 36 to get -44.

$$2x^{2}+12x-44=16\left(2-\left(x+6\right)\right)$$

To find the opposite of x+6, find the opposite of each term.

$$2x^{2}+12x-44=16\left(2-x-6\right)$$

Subtract 6 from 2 to get -4.

$$2x^{2}+12x-44=16\left(-4-x\right)$$

Use the distributive property to multiply 16 by -4-x.

$$2x^{2}+12x-44=-64-16x$$

Subtract -64 from both sides.

$$2x^{2}+12x-44-\left(-64\right)=-16x$$

The opposite of -64 is 64.

$$2x^{2}+12x-44+64=-16x$$

Add 16x to both sides.

$$2x^{2}+12x-44+64+16x=0$$

Add -44 and 64 to get 20.

$$2x^{2}+12x+20+16x=0$$

Combine 12x and 16x to get 28x.

$$2x^{2}+28x+20=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, 28 for b, and 20 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-28±\sqrt{28^{2}-4\times 2\times 20}}{2\times 2}$$

Square 28.

$$x=\frac{-28±\sqrt{784-4\times 2\times 20}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-28±\sqrt{784-8\times 20}}{2\times 2}$$

Multiply -8 times 20.

$$x=\frac{-28±\sqrt{784-160}}{2\times 2}$$

Add 784 to -160.

$$x=\frac{-28±\sqrt{624}}{2\times 2}$$

Take the square root of 624.

$$x=\frac{-28±4\sqrt{39}}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{-28±4\sqrt{39}}{4}$$

Now solve the equation x=\frac{-28±4\sqrt{39}}{4} when ± is plus. Add -28 to 4\sqrt{39}.

$$x=\frac{4\sqrt{39}-28}{4}$$

Divide -28+4\sqrt{39} by 4.

$$x=\sqrt{39}-7$$

Now solve the equation x=\frac{-28±4\sqrt{39}}{4} when ± is minus. Subtract 4\sqrt{39} from -28.

$$x=\frac{-4\sqrt{39}-28}{4}$$

Divide -28-4\sqrt{39} by 4.

$$x=-\sqrt{39}-7$$

The equation is now solved.

$$x=\sqrt{39}-7 x=-\sqrt{39}-7$$

Steps for Completing the Square

Multiply both sides of the equation by 4.

$$x^{2}+\left(x+6\right)^{2}-80=16\left(2-\left(x+6\right)\right)$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+6\right)^{2}.

$$x^{2}+x^{2}+12x+36-80=16\left(2-\left(x+6\right)\right)$$

Combine x^{2} and x^{2} to get 2x^{2}.

$$2x^{2}+12x+36-80=16\left(2-\left(x+6\right)\right)$$

Subtract 80 from 36 to get -44.

$$2x^{2}+12x-44=16\left(2-\left(x+6\right)\right)$$

To find the opposite of x+6, find the opposite of each term.

$$2x^{2}+12x-44=16\left(2-x-6\right)$$

Subtract 6 from 2 to get -4.

$$2x^{2}+12x-44=16\left(-4-x\right)$$

Use the distributive property to multiply 16 by -4-x.

$$2x^{2}+12x-44=-64-16x$$

Add 16x to both sides.

$$2x^{2}+12x-44+16x=-64$$

Combine 12x and 16x to get 28x.

$$2x^{2}+28x-44=-64$$

Add 44 to both sides.

$$2x^{2}+28x=-64+44$$

Add -64 and 44 to get -20.

$$2x^{2}+28x=-20$$

Divide both sides by 2.

$$\frac{2x^{2}+28x}{2}=-\frac{20}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}+\frac{28}{2}x=-\frac{20}{2}$$

Divide 28 by 2.

$$x^{2}+14x=-\frac{20}{2}$$

Divide -20 by 2.

$$x^{2}+14x=-10$$

Divide 14, the coefficient of the x term, by 2 to get 7. Then add the square of 7 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+14x+7^{2}=-10+7^{2}$$

Square 7.

$$x^{2}+14x+49=-10+49$$

Add -10 to 49.

$$x^{2}+14x+49=39$$

Factor x^{2}+14x+49. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+7\right)^{2}=39$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+7\right)^{2}}=\sqrt{39}$$

Simplify.

$$x+7=\sqrt{39} x+7=-\sqrt{39}$$

Subtract 7 from both sides of the equation.

$$x=\sqrt{39}-7 x=-\sqrt{39}-7$$