Solve -5x^2-12=-32

$$-5 { x }^{ 2 } -12 = -32$$

Solve for x

$$x=2$$

$$x=-2$$

Steps by Finding Square Root

Add 12 to both sides.

$$-5x^{2}=-32+12$$

Add -32 and 12 to get -20.

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

Divide both sides by -5.

$$x^{2}=\frac{-20}{-5}$$

Divide -20 by -5 to get 4.

$$x^{2}=4$$

Take the square root of both sides of the equation.

$$x=2 x=-2$$

Steps Using the Quadratic Formula

Add 32 to both sides.

$$-5x^{2}-12+32=0$$

Add -12 and 32 to get 20.

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

This equation is in standard form: ax^{2}+bx+c=0. Substitute -5 for a, 0 for b, and 20 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{0±\sqrt{0^{2}-4\left(-5\right)\times 20}}{2\left(-5\right)}$$

Square 0.

$$x=\frac{0±\sqrt{-4\left(-5\right)\times 20}}{2\left(-5\right)}$$

Multiply -4 times -5.

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

Multiply 20 times 20.

$$x=\frac{0±\sqrt{400}}{2\left(-5\right)}$$

Take the square root of 400.

$$x=\frac{0±20}{2\left(-5\right)}$$

Multiply 2 times -5.

$$x=\frac{0±20}{-10}$$

Now solve the equation x=\frac{0±20}{-10} when ± is plus. Divide 20 by -10.

$$x=-2$$

Now solve the equation x=\frac{0±20}{-10} when ± is minus. Divide -20 by -10.

$$x=2$$

The equation is now solved.

$$x=-2 x=2$$