Solve 20x=64-2(x^2

$$20x=64-2( { x }^{ 2 }$$

Solve for x (complex solution)

$$x=\sqrt{57}-5\approx 2.549834435$$

$$x=-\left(\sqrt{57}+5\right)\approx -12.549834435$$

Steps Using the Quadratic Formula

Subtract 64 from both sides.

$$20x-64=-2x^{2}$$

Add 2x^{2} to both sides.

$$20x-64+2x^{2}=0$$

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.

$$2x^{2}+20x-64=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, 20 for b, and -64 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-20±\sqrt{20^{2}-4\times 2\left(-64\right)}}{2\times 2}$$

Square 20.

$$x=\frac{-20±\sqrt{400-4\times 2\left(-64\right)}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-20±\sqrt{400-8\left(-64\right)}}{2\times 2}$$

Multiply -8 times -64.

$$x=\frac{-20±\sqrt{400+512}}{2\times 2}$$

Add 400 to 512.

$$x=\frac{-20±\sqrt{912}}{2\times 2}$$

Take the square root of 912.

$$x=\frac{-20±4\sqrt{57}}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{-20±4\sqrt{57}}{4}$$

Now solve the equation x=\frac{-20±4\sqrt{57}}{4} when ± is plus. Add -20 to 4\sqrt{57}.

$$x=\frac{4\sqrt{57}-20}{4}$$

Divide -20+4\sqrt{57} by 4.

$$x=\sqrt{57}-5$$

Now solve the equation x=\frac{-20±4\sqrt{57}}{4} when ± is minus. Subtract 4\sqrt{57} from -20.

$$x=\frac{-4\sqrt{57}-20}{4}$$

Divide -20-4\sqrt{57} by 4.

$$x=-\sqrt{57}-5$$

The equation is now solved.

$$x=\sqrt{57}-5 x=-\sqrt{57}-5$$

Steps for Completing the Square

Add 2x^{2} to both sides.

$$20x+2x^{2}=64$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$2x^{2}+20x=64$$

Divide both sides by 2.

$$\frac{2x^{2}+20x}{2}=\frac{64}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}+\frac{20}{2}x=\frac{64}{2}$$

Divide 20 by 2.

$$x^{2}+10x=\frac{64}{2}$$

Divide 64 by 2.

$$x^{2}+10x=32$$

Divide 10, the coefficient of the x term, by 2 to get 5. Then add the square of 5 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+10x+5^{2}=32+5^{2}$$

Square 5.

$$x^{2}+10x+25=32+25$$

Add 32 to 25.

$$x^{2}+10x+25=57$$

Factor x^{2}+10x+25. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+5\right)^{2}=57$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+5\right)^{2}}=\sqrt{57}$$

Simplify.

$$x+5=\sqrt{57} x+5=-\sqrt{57}$$

Subtract 5 from both sides of the equation.

$$x=\sqrt{57}-5 x=-\sqrt{57}-5$$

Solve for x

$$x=\sqrt{57}-5\approx 2.549834435$$

$$x=-\sqrt{57}-5\approx -12.549834435$$

Steps Using the Quadratic Formula

Subtract 64 from both sides.

$$20x-64=-2x^{2}$$

Add 2x^{2} to both sides.

$$20x-64+2x^{2}=0$$

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.

$$2x^{2}+20x-64=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, 20 for b, and -64 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-20±\sqrt{20^{2}-4\times 2\left(-64\right)}}{2\times 2}$$

Square 20.

$$x=\frac{-20±\sqrt{400-4\times 2\left(-64\right)}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-20±\sqrt{400-8\left(-64\right)}}{2\times 2}$$

Multiply -8 times -64.

$$x=\frac{-20±\sqrt{400+512}}{2\times 2}$$

Add 400 to 512.

$$x=\frac{-20±\sqrt{912}}{2\times 2}$$

Take the square root of 912.

$$x=\frac{-20±4\sqrt{57}}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{-20±4\sqrt{57}}{4}$$

Now solve the equation x=\frac{-20±4\sqrt{57}}{4} when ± is plus. Add -20 to 4\sqrt{57}.

$$x=\frac{4\sqrt{57}-20}{4}$$

Divide -20+4\sqrt{57} by 4.

$$x=\sqrt{57}-5$$

Now solve the equation x=\frac{-20±4\sqrt{57}}{4} when ± is minus. Subtract 4\sqrt{57} from -20.

$$x=\frac{-4\sqrt{57}-20}{4}$$

Divide -20-4\sqrt{57} by 4.

$$x=-\sqrt{57}-5$$

The equation is now solved.

$$x=\sqrt{57}-5 x=-\sqrt{57}-5$$

Steps for Completing the Square

Add 2x^{2} to both sides.

$$20x+2x^{2}=64$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$2x^{2}+20x=64$$

Divide both sides by 2.

$$\frac{2x^{2}+20x}{2}=\frac{64}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}+\frac{20}{2}x=\frac{64}{2}$$

Divide 20 by 2.

$$x^{2}+10x=\frac{64}{2}$$

Divide 64 by 2.

$$x^{2}+10x=32$$

Divide 10, the coefficient of the x term, by 2 to get 5. Then add the square of 5 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+10x+5^{2}=32+5^{2}$$

Square 5.

$$x^{2}+10x+25=32+25$$

Add 32 to 25.

$$x^{2}+10x+25=57$$

Factor x^{2}+10x+25. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+5\right)^{2}=57$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+5\right)^{2}}=\sqrt{57}$$

Simplify.

$$x+5=\sqrt{57} x+5=-\sqrt{57}$$

Subtract 5 from both sides of the equation.

$$x=\sqrt{57}-5 x=-\sqrt{57}-5$$