Solve t^2=1200

$$t ^ { 2 } = 1200$$

Solve for t

$$t=20\sqrt{3}\approx 34.641016151$$

$$t=-20\sqrt{3}\approx -34.641016151$$

Steps by Finding Square Root

Take the square root of both sides of the equation.

$$t=20\sqrt{3} t=-20\sqrt{3}$$

Steps Using the Quadratic Formula

Subtract 1200 from both sides.

$$t^{2}-1200=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 0 for b, and -1200 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$t=\frac{0±\sqrt{0^{2}-4\left(-1200\right)}}{2}$$

Square 0.

$$t=\frac{0±\sqrt{-4\left(-1200\right)}}{2}$$

Multiply -4 times -1200.

$$t=\frac{0±\sqrt{4800}}{2}$$

Take the square root of 4800.

$$t=\frac{0±40\sqrt{3}}{2}$$

Now solve the equation t=\frac{0±40\sqrt{3}}{2} when ± is plus.

$$t=20\sqrt{3}$$

Now solve the equation t=\frac{0±40\sqrt{3}}{2} when ± is minus.

$$t=-20\sqrt{3}$$

The equation is now solved.

$$t=20\sqrt{3} t=-20\sqrt{3}$$