Solve 180=1/2*10*t^2

$$180 = \frac { 1 } { 2 } \cdot 10 \cdot t ^ { 2 }$$

Solve for t

$$t=6$$

$$t=-6$$

Steps Using Difference of Squares

Multiply \frac{1}{2} and 10 to get 5.

$$180=5t^{2}$$

Swap sides so that all variable terms are on the left hand side.

$$5t^{2}=180$$

Subtract 180 from both sides.

$$5t^{2}-180=0$$

Divide both sides by 5.

$$t^{2}-36=0$$

Consider t^{2}-36. Rewrite t^{2}-36 as t^{2}-6^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

$$\left(t-6\right)\left(t+6\right)=0$$

To find equation solutions, solve t-6=0 and t+6=0.

$$t=6 t=-6$$

Steps by Finding Square Root

Multiply \frac{1}{2} and 10 to get 5.

$$180=5t^{2}$$

Swap sides so that all variable terms are on the left hand side.

$$5t^{2}=180$$

Divide both sides by 5.

$$t^{2}=\frac{180}{5}$$

Divide 180 by 5 to get 36.

$$t^{2}=36$$

Take the square root of both sides of the equation.

$$t=6 t=-6$$

Steps Using the Quadratic Formula

Multiply \frac{1}{2} and 10 to get 5.

$$180=5t^{2}$$

Swap sides so that all variable terms are on the left hand side.

$$5t^{2}=180$$

Subtract 180 from both sides.

$$5t^{2}-180=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 5 for a, 0 for b, and -180 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$t=\frac{0±\sqrt{0^{2}-4\times 5\left(-180\right)}}{2\times 5}$$

Square 0.

$$t=\frac{0±\sqrt{-4\times 5\left(-180\right)}}{2\times 5}$$

Multiply -4 times 5.

$$t=\frac{0±\sqrt{-20\left(-180\right)}}{2\times 5}$$

Multiply -20 times -180.

$$t=\frac{0±\sqrt{3600}}{2\times 5}$$

Take the square root of 3600.

$$t=\frac{0±60}{2\times 5}$$

Multiply 2 times 5.

$$t=\frac{0±60}{10}$$

Now solve the equation t=\frac{0±60}{10} when ± is plus. Divide 60 by 10.

$$t=6$$

Now solve the equation t=\frac{0±60}{10} when ± is minus. Divide -60 by 10.

$$t=-6$$

The equation is now solved.

$$t=6 t=-6$$