Solve x^2+left(2x+2right)^2=left(28-3xright)^2

$${ x }^{ 2 } + { \left(2x+2 \right) }^{ 2 } = { \left(28-3x \right) }^{ 2 }$$

Solve for x

$$x=5$$

$$x=39$$

Steps Using the Quadratic Formula

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

$$x^{2}+4x^{2}+8x+4=\left(28-3x\right)^{2}$$

Combine x^{2} and 4x^{2} to get 5x^{2}.

$$5x^{2}+8x+4=\left(28-3x\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(28-3x\right)^{2}.

$$5x^{2}+8x+4=784-168x+9x^{2}$$

Subtract 784 from both sides.

$$5x^{2}+8x+4-784=-168x+9x^{2}$$

Subtract 784 from 4 to get -780.

$$5x^{2}+8x-780=-168x+9x^{2}$$

Add 168x to both sides.

$$5x^{2}+8x-780+168x=9x^{2}$$

Combine 8x and 168x to get 176x.

$$5x^{2}+176x-780=9x^{2}$$

Subtract 9x^{2} from both sides.

$$5x^{2}+176x-780-9x^{2}=0$$

Combine 5x^{2} and -9x^{2} to get -4x^{2}.

$$-4x^{2}+176x-780=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -4 for a, 176 for b, and -780 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-176±\sqrt{176^{2}-4\left(-4\right)\left(-780\right)}}{2\left(-4\right)}$$

Square 176.

$$x=\frac{-176±\sqrt{30976-4\left(-4\right)\left(-780\right)}}{2\left(-4\right)}$$

Multiply -4 times -4.

$$x=\frac{-176±\sqrt{30976+16\left(-780\right)}}{2\left(-4\right)}$$

Multiply 16 times -780.

$$x=\frac{-176±\sqrt{30976-12480}}{2\left(-4\right)}$$

Add 30976 to -12480.

$$x=\frac{-176±\sqrt{18496}}{2\left(-4\right)}$$

Take the square root of 18496.

$$x=\frac{-176±136}{2\left(-4\right)}$$

Multiply 2 times -4.

$$x=\frac{-176±136}{-8}$$

Now solve the equation x=\frac{-176±136}{-8} when ± is plus. Add -176 to 136.

$$x=-\frac{40}{-8}$$

Divide -40 by -8.

$$x=5$$

Now solve the equation x=\frac{-176±136}{-8} when ± is minus. Subtract 136 from -176.

$$x=-\frac{312}{-8}$$

Divide -312 by -8.

$$x=39$$

The equation is now solved.

$$x=5 x=39$$

Steps for Completing the Square

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

$$x^{2}+4x^{2}+8x+4=\left(28-3x\right)^{2}$$

Combine x^{2} and 4x^{2} to get 5x^{2}.

$$5x^{2}+8x+4=\left(28-3x\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(28-3x\right)^{2}.

$$5x^{2}+8x+4=784-168x+9x^{2}$$

Add 168x to both sides.

$$5x^{2}+8x+4+168x=784+9x^{2}$$

Combine 8x and 168x to get 176x.

$$5x^{2}+176x+4=784+9x^{2}$$

Subtract 9x^{2} from both sides.

$$5x^{2}+176x+4-9x^{2}=784$$

Combine 5x^{2} and -9x^{2} to get -4x^{2}.

$$-4x^{2}+176x+4=784$$

Subtract 4 from both sides.

$$-4x^{2}+176x=784-4$$

Subtract 4 from 784 to get 780.

$$-4x^{2}+176x=780$$

Divide both sides by -4.

$$\frac{-4x^{2}+176x}{-4}=\frac{780}{-4}$$

Dividing by -4 undoes the multiplication by -4.

$$x^{2}+\frac{176}{-4}x=\frac{780}{-4}$$

Divide 176 by -4.

$$x^{2}-44x=\frac{780}{-4}$$

Divide 780 by -4.

$$x^{2}-44x=-195$$

Divide -44, the coefficient of the x term, by 2 to get -22. Then add the square of -22 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-44x+\left(-22\right)^{2}=-195+\left(-22\right)^{2}$$

Square -22.

$$x^{2}-44x+484=-195+484$$

Add -195 to 484.

$$x^{2}-44x+484=289$$

Factor x^{2}-44x+484. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-22\right)^{2}=289$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-22\right)^{2}}=\sqrt{289}$$

Simplify.

$$x-22=17 x-22=-17$$

Add 22 to both sides of the equation.

$$x=39 x=5$$