Solve x^2-10x+25=-5

$$x ^ { 2 } - 10 x + 25 = - 5$$

Solve for x (complex solution)

$$x=5+\sqrt{5}i\approx 5+2.236067977i$$

$$x=-\sqrt{5}i+5\approx 5-2.236067977i$$

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.

$$x^{2}-10x+25=-5$$

Add 5 to both sides of the equation.

$$x^{2}-10x+25-\left(-5\right)=-5-\left(-5\right)$$

Subtracting -5 from itself leaves 0.

$$x^{2}-10x+25-\left(-5\right)=0$$

Subtract -5 from 25.

$$x^{2}-10x+30=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -10 for b, and 30 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-10\right)±\sqrt{\left(-10\right)^{2}-4\times 30}}{2}$$

Square -10.

$$x=\frac{-\left(-10\right)±\sqrt{100-4\times 30}}{2}$$

Multiply -4 times 30.

$$x=\frac{-\left(-10\right)±\sqrt{100-120}}{2}$$

Add 100 to -120.

$$x=\frac{-\left(-10\right)±\sqrt{-20}}{2}$$

Take the square root of -20.

$$x=\frac{-\left(-10\right)±2\sqrt{5}i}{2}$$

The opposite of -10 is 10.

$$x=\frac{10±2\sqrt{5}i}{2}$$

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

$$x=\frac{10+2\sqrt{5}i}{2}$$

Divide 10+2i\sqrt{5} by 2.

$$x=5+\sqrt{5}i$$

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

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

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

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

The equation is now solved.

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

Steps for Completing the Square

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.

$$x^{2}-10x+25=-5$$

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}.

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

Take the square root of both sides of the equation.

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

Simplify.

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

Add 5 to both sides of the equation.

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