Solve 16+x^2=2*(4-x)(4-x)

$$16+ { x }^{ 2 } =2 \times (4-x)(4-x)$$

Solve for x

$$x=4\sqrt{3}+8\approx 14.92820323$$

$$x=8-4\sqrt{3}\approx 1.07179677$$

Steps Using the Quadratic Formula

Multiply 4-x and 4-x to get \left(4-x\right)^{2}.

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

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(4-x\right)^{2}.

$$16+x^{2}=2\left(16-8x+x^{2}\right)$$

Use the distributive property to multiply 2 by 16-8x+x^{2}.

$$16+x^{2}=32-16x+2x^{2}$$

Subtract 32 from both sides.

$$16+x^{2}-32=-16x+2x^{2}$$

Subtract 32 from 16 to get -16.

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

Add 16x to both sides.

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

Subtract 2x^{2} from both sides.

$$-16+x^{2}+16x-2x^{2}=0$$

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

$$-16-x^{2}+16x=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$-x^{2}+16x-16=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, 16 for b, and -16 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-16±\sqrt{16^{2}-4\left(-1\right)\left(-16\right)}}{2\left(-1\right)}$$

Square 16.

$$x=\frac{-16±\sqrt{256-4\left(-1\right)\left(-16\right)}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-16±\sqrt{256+4\left(-16\right)}}{2\left(-1\right)}$$

Multiply 4 times -16.

$$x=\frac{-16±\sqrt{256-64}}{2\left(-1\right)}$$

Add 256 to -64.

$$x=\frac{-16±\sqrt{192}}{2\left(-1\right)}$$

Take the square root of 192.

$$x=\frac{-16±8\sqrt{3}}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{-16±8\sqrt{3}}{-2}$$

Now solve the equation x=\frac{-16±8\sqrt{3}}{-2} when ± is plus. Add -16 to 8\sqrt{3}.

$$x=\frac{8\sqrt{3}-16}{-2}$$

Divide -16+8\sqrt{3} by -2.

$$x=8-4\sqrt{3}$$

Now solve the equation x=\frac{-16±8\sqrt{3}}{-2} when ± is minus. Subtract 8\sqrt{3} from -16.

$$x=\frac{-8\sqrt{3}-16}{-2}$$

Divide -16-8\sqrt{3} by -2.

$$x=4\sqrt{3}+8$$

The equation is now solved.

$$x=8-4\sqrt{3} x=4\sqrt{3}+8$$

Steps for Completing the Square

Multiply 4-x and 4-x to get \left(4-x\right)^{2}.

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

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(4-x\right)^{2}.

$$16+x^{2}=2\left(16-8x+x^{2}\right)$$

Use the distributive property to multiply 2 by 16-8x+x^{2}.

$$16+x^{2}=32-16x+2x^{2}$$

Add 16x to both sides.

$$16+x^{2}+16x=32+2x^{2}$$

Subtract 2x^{2} from both sides.

$$16+x^{2}+16x-2x^{2}=32$$

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

$$16-x^{2}+16x=32$$

Subtract 16 from both sides.

$$-x^{2}+16x=32-16$$

Subtract 16 from 32 to get 16.

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

Divide both sides by -1.

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

Dividing by -1 undoes the multiplication by -1.

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

Divide 16 by -1.

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

Divide 16 by -1.

$$x^{2}-16x=-16$$

Divide -16, the coefficient of the x term, by 2 to get -8. Then add the square of -8 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-16x+\left(-8\right)^{2}=-16+\left(-8\right)^{2}$$

Square -8.

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

Add -16 to 64.

$$x^{2}-16x+64=48$$

Factor x^{2}-16x+64. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-8\right)^{2}=48$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-8\right)^{2}}=\sqrt{48}$$

Simplify.

$$x-8=4\sqrt{3} x-8=-4\sqrt{3}$$

Add 8 to both sides of the equation.

$$x=4\sqrt{3}+8 x=8-4\sqrt{3}$$