Solve x^2+10x+25=500

$$x ^ { 2 } + 10 x + 25 = 500$$

Solve for x

$$x=10\sqrt{5}-5\approx 17.360679775$$

$$x=-10\sqrt{5}-5\approx -27.360679775$$

Steps Using the Quadratic Formula

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.

$$x^{2}+10x+25=500$$

Subtract 500 from both sides of the equation.

$$x^{2}+10x+25-500=500-500$$

Subtracting 500 from itself leaves 0.

$$x^{2}+10x+25-500=0$$

Subtract 500 from 25.

$$x^{2}+10x-475=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 10 for b, and -475 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-10±\sqrt{10^{2}-4\left(-475\right)}}{2}$$

Square 10.

$$x=\frac{-10±\sqrt{100-4\left(-475\right)}}{2}$$

Multiply -4 times -475.

$$x=\frac{-10±\sqrt{100+1900}}{2}$$

Add 100 to 1900.

$$x=\frac{-10±\sqrt{2000}}{2}$$

Take the square root of 2000.

$$x=\frac{-10±20\sqrt{5}}{2}$$

Now solve the equation x=\frac{-10±20\sqrt{5}}{2} when ± is plus. Add -10 to 20\sqrt{5}.

$$x=\frac{20\sqrt{5}-10}{2}$$

Divide -10+20\sqrt{5} by 2.

$$x=10\sqrt{5}-5$$

Now solve the equation x=\frac{-10±20\sqrt{5}}{2} when ± is minus. Subtract 20\sqrt{5} from -10.

$$x=\frac{-20\sqrt{5}-10}{2}$$

Divide -10-20\sqrt{5} by 2.

$$x=-10\sqrt{5}-5$$

The equation is now solved.

$$x=10\sqrt{5}-5 x=-10\sqrt{5}-5$$

Steps for Completing the Square

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}.

$$\left(x+5\right)^{2}=500$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+5\right)^{2}}=\sqrt{500}$$

Simplify.

$$x+5=10\sqrt{5} x+5=-10\sqrt{5}$$

Subtract 5 from both sides of the equation.

$$x=10\sqrt{5}-5 x=-10\sqrt{5}-5$$