Solve x^2+10x+25=7

$$x ^ { 2 } + 10 x + 25 = 7$$

Solve for x (complex solution)

$$x=\sqrt{7}-5\approx -2.354248689$$

$$x=-\left(\sqrt{7}+5\right)\approx -7.645751311$$

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.

$$x^{2}+10x+25=7$$

Subtract 7 from both sides of the equation.

$$x^{2}+10x+25-7=7-7$$

Subtracting 7 from itself leaves 0.

$$x^{2}+10x+25-7=0$$

Subtract 7 from 25.

$$x^{2}+10x+18=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 10 for b, and 18 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-10±\sqrt{10^{2}-4\times 18}}{2}$$

Square 10.

$$x=\frac{-10±\sqrt{100-4\times 18}}{2}$$

Multiply -4 times 18.

$$x=\frac{-10±\sqrt{100-72}}{2}$$

Add 100 to -72.

$$x=\frac{-10±\sqrt{28}}{2}$$

Take the square root of 28.

$$x=\frac{-10±2\sqrt{7}}{2}$$

Now solve the equation x=\frac{-10±2\sqrt{7}}{2} when ± is plus. Add -10 to 2\sqrt{7}.

$$x=\frac{2\sqrt{7}-10}{2}$$

Divide -10+2\sqrt{7} by 2.

$$x=\sqrt{7}-5$$

Now solve the equation x=\frac{-10±2\sqrt{7}}{2} when ± is minus. Subtract 2\sqrt{7} from -10.

$$x=\frac{-2\sqrt{7}-10}{2}$$

Divide -10-2\sqrt{7} by 2.

$$x=-\sqrt{7}-5$$

The equation is now solved.

$$x=\sqrt{7}-5 x=-\sqrt{7}-5$$

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}.

$$\left(x+5\right)^{2}=7$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+5\right)^{2}}=\sqrt{7}$$

Simplify.

$$x+5=\sqrt{7} x+5=-\sqrt{7}$$

Subtract 5 from both sides of the equation.

$$x=\sqrt{7}-5 x=-\sqrt{7}-5$$

Solve for x

$$x=\sqrt{7}-5\approx -2.354248689$$

$$x=-\sqrt{7}-5\approx -7.645751311$$

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.

$$x^{2}+10x+25=7$$

Subtract 7 from both sides of the equation.

$$x^{2}+10x+25-7=7-7$$

Subtracting 7 from itself leaves 0.

$$x^{2}+10x+25-7=0$$

Subtract 7 from 25.

$$x^{2}+10x+18=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 10 for b, and 18 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-10±\sqrt{10^{2}-4\times 18}}{2}$$

Square 10.

$$x=\frac{-10±\sqrt{100-4\times 18}}{2}$$

Multiply -4 times 18.

$$x=\frac{-10±\sqrt{100-72}}{2}$$

Add 100 to -72.

$$x=\frac{-10±\sqrt{28}}{2}$$

Take the square root of 28.

$$x=\frac{-10±2\sqrt{7}}{2}$$

Now solve the equation x=\frac{-10±2\sqrt{7}}{2} when ± is plus. Add -10 to 2\sqrt{7}.

$$x=\frac{2\sqrt{7}-10}{2}$$

Divide -10+2\sqrt{7} by 2.

$$x=\sqrt{7}-5$$

Now solve the equation x=\frac{-10±2\sqrt{7}}{2} when ± is minus. Subtract 2\sqrt{7} from -10.

$$x=\frac{-2\sqrt{7}-10}{2}$$

Divide -10-2\sqrt{7} by 2.

$$x=-\sqrt{7}-5$$

The equation is now solved.

$$x=\sqrt{7}-5 x=-\sqrt{7}-5$$

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}.

$$\left(x+5\right)^{2}=7$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+5\right)^{2}}=\sqrt{7}$$

Simplify.

$$x+5=\sqrt{7} x+5=-\sqrt{7}$$

Subtract 5 from both sides of the equation.

$$x=\sqrt{7}-5 x=-\sqrt{7}-5$$