Solve x=x^2+10x+25

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

Solve for x (complex solution)

$$x=\frac{-\sqrt{19}i-9}{2}\approx -4.5-2.179449472i$$

$$x=\frac{-9+\sqrt{19}i}{2}\approx -4.5+2.179449472i$$

Steps Using the Quadratic Formula

Subtract x^{2} from both sides.

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

Subtract 10x from both sides.

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

Combine x and -10x to get -9x.

$$-9x-x^{2}=25$$

Subtract 25 from both sides.

$$-9x-x^{2}-25=0$$

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.

$$-x^{2}-9x-25=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, -9 for b, and -25 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-9\right)±\sqrt{\left(-9\right)^{2}-4\left(-1\right)\left(-25\right)}}{2\left(-1\right)}$$

Square -9.

$$x=\frac{-\left(-9\right)±\sqrt{81-4\left(-1\right)\left(-25\right)}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-\left(-9\right)±\sqrt{81+4\left(-25\right)}}{2\left(-1\right)}$$

Multiply 4 times -25.

$$x=\frac{-\left(-9\right)±\sqrt{81-100}}{2\left(-1\right)}$$

Add 81 to -100.

$$x=\frac{-\left(-9\right)±\sqrt{-19}}{2\left(-1\right)}$$

Take the square root of -19.

$$x=\frac{-\left(-9\right)±\sqrt{19}i}{2\left(-1\right)}$$

The opposite of -9 is 9.

$$x=\frac{9±\sqrt{19}i}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{9±\sqrt{19}i}{-2}$$

Now solve the equation x=\frac{9±\sqrt{19}i}{-2} when ± is plus. Add 9 to i\sqrt{19}.

$$x=\frac{9+\sqrt{19}i}{-2}$$

Divide 9+i\sqrt{19} by -2.

$$x=\frac{-\sqrt{19}i-9}{2}$$

Now solve the equation x=\frac{9±\sqrt{19}i}{-2} when ± is minus. Subtract i\sqrt{19} from 9.

$$x=\frac{-\sqrt{19}i+9}{-2}$$

Divide 9-i\sqrt{19} by -2.

$$x=\frac{-9+\sqrt{19}i}{2}$$

The equation is now solved.

$$x=\frac{-\sqrt{19}i-9}{2} x=\frac{-9+\sqrt{19}i}{2}$$

Steps for Completing the Square

Subtract x^{2} from both sides.

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

Subtract 10x from both sides.

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

Combine x and -10x to get -9x.

$$-9x-x^{2}=25$$

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.

$$-x^{2}-9x=25$$

Divide both sides by -1.

$$\frac{-x^{2}-9x}{-1}=\frac{25}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$x^{2}+\left(-\frac{9}{-1}\right)x=\frac{25}{-1}$$

Divide -9 by -1.

$$x^{2}+9x=\frac{25}{-1}$$

Divide 25 by -1.

$$x^{2}+9x=-25$$

Divide 9, the coefficient of the x term, by 2 to get \frac{9}{2}. Then add the square of \frac{9}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+9x+\left(\frac{9}{2}\right)^{2}=-25+\left(\frac{9}{2}\right)^{2}$$

Square \frac{9}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+9x+\frac{81}{4}=-25+\frac{81}{4}$$

Add -25 to \frac{81}{4}.

$$x^{2}+9x+\frac{81}{4}=-\frac{19}{4}$$

Factor x^{2}+9x+\frac{81}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{9}{2}\right)^{2}=-\frac{19}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{9}{2}\right)^{2}}=\sqrt{-\frac{19}{4}}$$

Simplify.

$$x+\frac{9}{2}=\frac{\sqrt{19}i}{2} x+\frac{9}{2}=-\frac{\sqrt{19}i}{2}$$

Subtract \frac{9}{2} from both sides of the equation.

$$x=\frac{-9+\sqrt{19}i}{2} x=\frac{-\sqrt{19}i-9}{2}$$