Solve -5x^2+20=15

$$- 5 x ^ { 2 } + 20 = 15$$

Solve for x

$$x=1$$

$$x=-1$$

Steps by Finding Square Root

Subtract 20 from both sides.

$$-5x^{2}=15-20$$

Subtract 20 from 15 to get -5.

$$-5x^{2}=-5$$

Divide both sides by -5.

$$x^{2}=\frac{-5}{-5}$$

Divide -5 by -5 to get 1.

$$x^{2}=1$$

Take the square root of both sides of the equation.

$$x=1 x=-1$$

Steps Using the Quadratic Formula

Subtract 15 from both sides.

$$-5x^{2}+20-15=0$$

Subtract 15 from 20 to get 5.

$$-5x^{2}+5=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -5 for a, 0 for b, and 5 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{0±\sqrt{0^{2}-4\left(-5\right)\times 5}}{2\left(-5\right)}$$

Square 0.

$$x=\frac{0±\sqrt{-4\left(-5\right)\times 5}}{2\left(-5\right)}$$

Multiply -4 times -5.

$$x=\frac{0±\sqrt{20\times 5}}{2\left(-5\right)}$$

Multiply 20 times 5.

$$x=\frac{0±\sqrt{100}}{2\left(-5\right)}$$

Take the square root of 100.

$$x=\frac{0±10}{2\left(-5\right)}$$

Multiply 2 times -5.

$$x=\frac{0±10}{-10}$$

Now solve the equation x=\frac{0±10}{-10} when ± is plus. Divide 10 by -10.

$$x=-1$$

Now solve the equation x=\frac{0±10}{-10} when ± is minus. Divide -10 by -10.

$$x=1$$

The equation is now solved.

$$x=-1 x=1$$