Solve -1-5x^2=-321

$$- 1 - 5 x ^ { 2 } = - 321$$

Solve for x

$$x=8$$

$$x=-8$$

Steps by Finding Square Root

Add 1 to both sides.

$$-5x^{2}=-321+1$$

Add -321 and 1 to get -320.

$$-5x^{2}=-320$$

Divide both sides by -5.

$$x^{2}=\frac{-320}{-5}$$

Divide -320 by -5 to get 64.

$$x^{2}=64$$

Take the square root of both sides of the equation.

$$x=8 x=-8$$

Steps Using the Quadratic Formula

Add 321 to both sides.

$$-1-5x^{2}+321=0$$

Add -1 and 321 to get 320.

$$320-5x^{2}=0$$

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.

$$-5x^{2}+320=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -5 for a, 0 for b, and 320 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{0±\sqrt{0^{2}-4\left(-5\right)\times 320}}{2\left(-5\right)}$$

Square 0.

$$x=\frac{0±\sqrt{-4\left(-5\right)\times 320}}{2\left(-5\right)}$$

Multiply -4 times -5.

$$x=\frac{0±\sqrt{20\times 320}}{2\left(-5\right)}$$

Multiply 20 times 320.

$$x=\frac{0±\sqrt{6400}}{2\left(-5\right)}$$

Take the square root of 6400.

$$x=\frac{0±80}{2\left(-5\right)}$$

Multiply 2 times -5.

$$x=\frac{0±80}{-10}$$

Now solve the equation x=\frac{0±80}{-10} when ± is plus. Divide 80 by -10.

$$x=-8$$

Now solve the equation x=\frac{0±80}{-10} when ± is minus. Divide -80 by -10.

$$x=8$$

The equation is now solved.

$$x=-8 x=8$$