Solve 35=2(x-5)^2

$$35 = 2 ( x - 5 ) ^ { 2 }$$

Solve for x

$$x=\frac{\sqrt{70}}{2}+5\approx 9.183300133$$

$$x=-\frac{\sqrt{70}}{2}+5\approx 0.816699867$$

Steps Using the Quadratic Formula

Divide both sides by 2.

$$\frac{35}{2}=\left(x-5\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-5\right)^{2}.

$$\frac{35}{2}=x^{2}-10x+25$$

Swap sides so that all variable terms are on the left hand side.

$$x^{2}-10x+25=\frac{35}{2}$$

Subtract \frac{35}{2} from both sides.

$$x^{2}-10x+25-\frac{35}{2}=0$$

Subtract \frac{35}{2} from 25 to get \frac{15}{2}.

$$x^{2}-10x+\frac{15}{2}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -10 for b, and \frac{15}{2} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-10\right)±\sqrt{\left(-10\right)^{2}-4\times \frac{15}{2}}}{2}$$

Square -10.

$$x=\frac{-\left(-10\right)±\sqrt{100-4\times \frac{15}{2}}}{2}$$

Multiply -4 times \frac{15}{2}.

$$x=\frac{-\left(-10\right)±\sqrt{100-30}}{2}$$

Add 100 to -30.

$$x=\frac{-\left(-10\right)±\sqrt{70}}{2}$$

The opposite of -10 is 10.

$$x=\frac{10±\sqrt{70}}{2}$$

Now solve the equation x=\frac{10±\sqrt{70}}{2} when ± is plus. Add 10 to \sqrt{70}.

$$x=\frac{\sqrt{70}+10}{2}$$

Divide 10+\sqrt{70} by 2.

$$x=\frac{\sqrt{70}}{2}+5$$

Now solve the equation x=\frac{10±\sqrt{70}}{2} when ± is minus. Subtract \sqrt{70} from 10.

$$x=\frac{10-\sqrt{70}}{2}$$

Divide 10-\sqrt{70} by 2.

$$x=-\frac{\sqrt{70}}{2}+5$$

The equation is now solved.

$$x=\frac{\sqrt{70}}{2}+5 x=-\frac{\sqrt{70}}{2}+5$$

Steps for Completing the Square

Divide both sides by 2.

$$\frac{35}{2}=\left(x-5\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-5\right)^{2}.

$$\frac{35}{2}=x^{2}-10x+25$$

Swap sides so that all variable terms are on the left hand side.

$$x^{2}-10x+25=\frac{35}{2}$$

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}.

$$\left(x-5\right)^{2}=\frac{35}{2}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-5\right)^{2}}=\sqrt{\frac{35}{2}}$$

Simplify.

$$x-5=\frac{\sqrt{70}}{2} x-5=-\frac{\sqrt{70}}{2}$$

Add 5 to both sides of the equation.

$$x=\frac{\sqrt{70}}{2}+5 x=-\frac{\sqrt{70}}{2}+5$$