Solve 180-3x*2x-20=0

$$180-3x \times 2x-20=0$$

Solve for x

$$x = \frac{4 \sqrt{15}}{3} \approx 5.163977795$$

$$x = -\frac{4 \sqrt{15}}{3} \approx -5.163977795$$

Steps by Finding Square Root

Multiply x and x to get x^{2}.

$$180-3x^{2}\times 2-20=0$$

Multiply 3 and 2 to get 6.

$$180-6x^{2}-20=0$$

Subtract 20 from 180 to get 160.

$$160-6x^{2}=0$$

Subtract 160 from both sides. Anything subtracted from zero gives its negation.

$$-6x^{2}=-160$$

Divide both sides by -6.

$$x^{2}=\frac{-160}{-6}$$

Reduce the fraction \frac{-160}{-6} to lowest terms by extracting and canceling out -2.

$$x^{2}=\frac{80}{3}$$

Take the square root of both sides of the equation.

$$x=\frac{4\sqrt{15}}{3} x=-\frac{4\sqrt{15}}{3}$$

Steps Using the Quadratic Formula

Multiply x and x to get x^{2}.

$$180-3x^{2}\times 2-20=0$$

Multiply 3 and 2 to get 6.

$$180-6x^{2}-20=0$$

Subtract 20 from 180 to get 160.

$$160-6x^{2}=0$$

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.

$$-6x^{2}+160=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -6 for a, 0 for b, and 160 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{0±\sqrt{0^{2}-4\left(-6\right)\times 160}}{2\left(-6\right)}$$

Square 0.

$$x=\frac{0±\sqrt{-4\left(-6\right)\times 160}}{2\left(-6\right)}$$

Multiply -4 times -6.

$$x=\frac{0±\sqrt{24\times 160}}{2\left(-6\right)}$$

Multiply 24 times 160.

$$x=\frac{0±\sqrt{3840}}{2\left(-6\right)}$$

Take the square root of 3840.

$$x=\frac{0±16\sqrt{15}}{2\left(-6\right)}$$

Multiply 2 times -6.

$$x=\frac{0±16\sqrt{15}}{-12}$$

Now solve the equation x=\frac{0±16\sqrt{15}}{-12} when ± is plus.

$$x=-\frac{4\sqrt{15}}{3}$$

Now solve the equation x=\frac{0±16\sqrt{15}}{-12} when ± is minus.

$$x=\frac{4\sqrt{15}}{3}$$

The equation is now solved.

$$x=-\frac{4\sqrt{15}}{3} x=\frac{4\sqrt{15}}{3}$$