Solve -4/3x^2+12=0

$$- \frac { 4 } { 3 } x ^ { 2 } + 12 = 0$$

Solve for x

$$x=3$$

$$x=-3$$

Steps by Finding Square Root

Subtract 12 from both sides. Anything subtracted from zero gives its negation.

$$-\frac{4}{3}x^{2}=-12$$

Multiply both sides by -\frac{3}{4}, the reciprocal of -\frac{4}{3}.

$$x^{2}=-12\left(-\frac{3}{4}\right)$$

Multiply -12 and -\frac{3}{4} to get 9.

$$x^{2}=9$$

Take the square root of both sides of the equation.

$$x=3 x=-3$$

Steps Using the Quadratic Formula

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.

$$-\frac{4}{3}x^{2}+12=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -\frac{4}{3} for a, 0 for b, and 12 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{0±\sqrt{0^{2}-4\left(-\frac{4}{3}\right)\times 12}}{2\left(-\frac{4}{3}\right)}$$

Square 0.

$$x=\frac{0±\sqrt{-4\left(-\frac{4}{3}\right)\times 12}}{2\left(-\frac{4}{3}\right)}$$

Multiply -4 times -\frac{4}{3}.

$$x=\frac{0±\sqrt{\frac{16}{3}\times 12}}{2\left(-\frac{4}{3}\right)}$$

Multiply \frac{16}{3} times 12.

$$x=\frac{0±\sqrt{64}}{2\left(-\frac{4}{3}\right)}$$

Take the square root of 64.

$$x=\frac{0±8}{2\left(-\frac{4}{3}\right)}$$

Multiply 2 times -\frac{4}{3}.

$$x=\frac{0±8}{-\frac{8}{3}}$$

Now solve the equation x=\frac{0±8}{-\frac{8}{3}} when ± is plus. Divide 8 by -\frac{8}{3} by multiplying 8 by the reciprocal of -\frac{8}{3}.

$$x=-3$$

Now solve the equation x=\frac{0±8}{-\frac{8}{3}} when ± is minus. Divide -8 by -\frac{8}{3} by multiplying -8 by the reciprocal of -\frac{8}{3}.

$$x=3$$

The equation is now solved.

$$x=-3 x=3$$