Solve 12-t^2=0

$$12 - t ^ { 2 } = 0$$

Solve for t

$$t=2\sqrt{3}\approx 3.464101615$$

$$t=-2\sqrt{3}\approx -3.464101615$$

Steps by Finding Square Root

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

$$-t^{2}=-12$$

Divide both sides by -1.

$$t^{2}=\frac{-12}{-1}$$

Fraction \frac{-12}{-1} can be simplified to 12 by removing the negative sign from both the numerator and the denominator.

$$t^{2}=12$$

Take the square root of both sides of the equation.

$$t=2\sqrt{3} t=-2\sqrt{3}$$

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.

$$-t^{2}+12=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, 0 for b, and 12 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$t=\frac{0±\sqrt{0^{2}-4\left(-1\right)\times 12}}{2\left(-1\right)}$$

Square 0.

$$t=\frac{0±\sqrt{-4\left(-1\right)\times 12}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$t=\frac{0±\sqrt{4\times 12}}{2\left(-1\right)}$$

Multiply 4 times 12.

$$t=\frac{0±\sqrt{48}}{2\left(-1\right)}$$

Take the square root of 48.

$$t=\frac{0±4\sqrt{3}}{2\left(-1\right)}$$

Multiply 2 times -1.

$$t=\frac{0±4\sqrt{3}}{-2}$$

Now solve the equation t=\frac{0±4\sqrt{3}}{-2} when ± is plus.

$$t=-2\sqrt{3}$$

Now solve the equation t=\frac{0±4\sqrt{3}}{-2} when ± is minus.

$$t=2\sqrt{3}$$

The equation is now solved.

$$t=-2\sqrt{3} t=2\sqrt{3}$$