Solve 20-4x^2=18

$$20-4 { x }^{ 2 } =18$$

Solve for x

$$x=\frac{\sqrt{2}}{2}\approx 0.707106781$$

$$x=-\frac{\sqrt{2}}{2}\approx -0.707106781$$

Steps by Finding Square Root

Subtract 20 from both sides.

$$-4x^{2}=18-20$$

Subtract 20 from 18 to get -2.

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

Divide both sides by -4.

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

Reduce the fraction \frac{-2}{-4} to lowest terms by extracting and canceling out -2.

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

Take the square root of both sides of the equation.

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

Steps Using the Quadratic Formula

Subtract 18 from both sides.

$$20-4x^{2}-18=0$$

Subtract 18 from 20 to get 2.

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

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.

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

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

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

Square 0.

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

Multiply -4 times -4.

$$x=\frac{0±\sqrt{16\times 2}}{2\left(-4\right)}$$

Multiply 16 times 2.

$$x=\frac{0±\sqrt{32}}{2\left(-4\right)}$$

Take the square root of 32.

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

Multiply 2 times -4.

$$x=\frac{0±4\sqrt{2}}{-8}$$

Now solve the equation x=\frac{0±4\sqrt{2}}{-8} when ± is plus.

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

Now solve the equation x=\frac{0±4\sqrt{2}}{-8} when ± is minus.

$$x=\frac{\sqrt{2}}{2}$$

The equation is now solved.

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