Solve 4x+2=x^2+4x-2

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

Solve for x

$$x=-2$$

$$x=2$$

Steps by Finding Square Root

Subtract x^{2} from both sides.

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

Subtract 4x from both sides.

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

Combine 4x and -4x to get 0.

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

Subtract 2 from both sides.

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

Subtract 2 from -2 to get -4.

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

Divide both sides by -1.

$$x^{2}=\frac{-4}{-1}$$

Fraction \frac{-4}{-1} can be simplified to 4 by removing the negative sign from both the numerator and the denominator.

$$x^{2}=4$$

Take the square root of both sides of the equation.

$$x=2 x=-2$$

Steps Using the Quadratic Formula

Subtract x^{2} from both sides.

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

Subtract 4x from both sides.

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

Combine 4x and -4x to get 0.

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

Add 2 to both sides.

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

Add 2 and 2 to get 4.

$$4-x^{2}=0$$

Quadratic equations like this one, with an x^{2} term but no x term, can still be solved using the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}, once they are put in standard form: ax^{2}+bx+c=0.

$$-x^{2}+4=0$$

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

$$x=\frac{0±\sqrt{0^{2}-4\left(-1\right)\times 4}}{2\left(-1\right)}$$

Square 0.

$$x=\frac{0±\sqrt{-4\left(-1\right)\times 4}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{0±\sqrt{4\times 4}}{2\left(-1\right)}$$

Multiply 4 times 4.

$$x=\frac{0±\sqrt{16}}{2\left(-1\right)}$$

Take the square root of 16.

$$x=\frac{0±4}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{0±4}{-2}$$

Now solve the equation x=\frac{0±4}{-2} when ± is plus. Divide 4 by -2.

$$x=-2$$

Now solve the equation x=\frac{0±4}{-2} when ± is minus. Divide -4 by -2.

$$x=2$$

The equation is now solved.

$$x=-2 x=2$$