Solve x=x^2+4x-6

$$x = x ^ { 2 } + 4 x - 6$$

Solve for x

$$x = \frac{\sqrt{33} - 3}{2} \approx 1.372281323$$

$$x=\frac{-\sqrt{33}-3}{2}\approx -4.372281323$$

Steps Using the Quadratic Formula

Subtract x^{2} from both sides.

$$x-x^{2}=4x-6$$

Subtract 4x from both sides.

$$x-x^{2}-4x=-6$$

Combine x and -4x to get -3x.

$$-3x-x^{2}=-6$$

Add 6 to both sides.

$$-3x-x^{2}+6=0$$

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.

$$-x^{2}-3x+6=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, -3 for b, and 6 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\left(-1\right)\times 6}}{2\left(-1\right)}$$

Square -3.

$$x=\frac{-\left(-3\right)±\sqrt{9-4\left(-1\right)\times 6}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-\left(-3\right)±\sqrt{9+4\times 6}}{2\left(-1\right)}$$

Multiply 4 times 6.

$$x=\frac{-\left(-3\right)±\sqrt{9+24}}{2\left(-1\right)}$$

Add 9 to 24.

$$x=\frac{-\left(-3\right)±\sqrt{33}}{2\left(-1\right)}$$

The opposite of -3 is 3.

$$x=\frac{3±\sqrt{33}}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{3±\sqrt{33}}{-2}$$

Now solve the equation x=\frac{3±\sqrt{33}}{-2} when ± is plus. Add 3 to \sqrt{33}.

$$x=\frac{\sqrt{33}+3}{-2}$$

Divide 3+\sqrt{33} by -2.

$$x=\frac{-\sqrt{33}-3}{2}$$

Now solve the equation x=\frac{3±\sqrt{33}}{-2} when ± is minus. Subtract \sqrt{33} from 3.

$$x=\frac{3-\sqrt{33}}{-2}$$

Divide 3-\sqrt{33} by -2.

$$x=\frac{\sqrt{33}-3}{2}$$

The equation is now solved.

$$x=\frac{-\sqrt{33}-3}{2} x=\frac{\sqrt{33}-3}{2}$$

Steps for Completing the Square

Subtract x^{2} from both sides.

$$x-x^{2}=4x-6$$

Subtract 4x from both sides.

$$x-x^{2}-4x=-6$$

Combine x and -4x to get -3x.

$$-3x-x^{2}=-6$$

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.

$$-x^{2}-3x=-6$$

Divide both sides by -1.

$$\frac{-x^{2}-3x}{-1}=-\frac{6}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$x^{2}+\left(-\frac{3}{-1}\right)x=-\frac{6}{-1}$$

Divide -3 by -1.

$$x^{2}+3x=-\frac{6}{-1}$$

Divide -6 by -1.

$$x^{2}+3x=6$$

Divide 3, the coefficient of the x term, by 2 to get \frac{3}{2}. Then add the square of \frac{3}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+3x+\left(\frac{3}{2}\right)^{2}=6+\left(\frac{3}{2}\right)^{2}$$

Square \frac{3}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+3x+\frac{9}{4}=6+\frac{9}{4}$$

Add 6 to \frac{9}{4}.

$$x^{2}+3x+\frac{9}{4}=\frac{33}{4}$$

Factor x^{2}+3x+\frac{9}{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}.

$$\left(x+\frac{3}{2}\right)^{2}=\frac{33}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{3}{2}\right)^{2}}=\sqrt{\frac{33}{4}}$$

Simplify.

$$x+\frac{3}{2}=\frac{\sqrt{33}}{2} x+\frac{3}{2}=-\frac{\sqrt{33}}{2}$$

Subtract \frac{3}{2} from both sides of the equation.

$$x=\frac{\sqrt{33}-3}{2} x=\frac{-\sqrt{33}-3}{2}$$