Solve 0.64=1/2*2*t^2

$$0.64 = \frac { 1 } { 2 } \cdot 2 \cdot t ^ { 2 }$$

Solve for t

$$t=0.8$$

$$t=-0.8$$

Steps Using Difference of Squares

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

$$0.64=t^{2}$$

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

$$t^{2}=0.64$$

Subtract 0.64 from both sides.

$$t^{2}-0.64=0$$

Consider t^{2}-0.64. Rewrite t^{2}-0.64 as t^{2}-\left(\frac{4}{5}\right)^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

$$\left(t-\frac{4}{5}\right)\left(t+\frac{4}{5}\right)=0$$

To find equation solutions, solve t-\frac{4}{5}=0 and t+\frac{4}{5}=0.

$$t=\frac{4}{5} t=-\frac{4}{5}$$

Steps by Finding Square Root

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

$$0.64=t^{2}$$

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

$$t^{2}=0.64$$

Take the square root of both sides of the equation.

$$t=\frac{4}{5} t=-\frac{4}{5}$$

Steps Using the Quadratic Formula

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

$$0.64=t^{2}$$

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

$$t^{2}=0.64$$

Subtract 0.64 from both sides.

$$t^{2}-0.64=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 0 for b, and -0.64 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$t=\frac{0±\sqrt{0^{2}-4\left(-0.64\right)}}{2}$$

Square 0.

$$t=\frac{0±\sqrt{-4\left(-0.64\right)}}{2}$$

Multiply -4 times -0.64.

$$t=\frac{0±\sqrt{2.56}}{2}$$

Take the square root of 2.56.

$$t=\frac{0±\frac{8}{5}}{2}$$

Now solve the equation t=\frac{0±\frac{8}{5}}{2} when ± is plus.

$$t=\frac{4}{5}$$

Now solve the equation t=\frac{0±\frac{8}{5}}{2} when ± is minus.

$$t=-\frac{4}{5}$$

The equation is now solved.

$$t=\frac{4}{5} t=-\frac{4}{5}$$