t=-5t^2+180

Solution for t=-5t^2+180 equation:

t=-5t^2+180

We move all terms to the left:

t-(-5t^2+180)=0

We get rid of parentheses

5t^2+t-180=0

a = 5; b = 1; c = -180;
Δ = b2-4ac
Δ = 12-4·5·(-180)
Δ = 3601
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-\sqrt{3601}}{2*5}=\frac{-1-\sqrt{3601}}{10}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{3601}}{2*5}=\frac{-1+\sqrt{3601}}{10}
$