t=-5t^2+10t+20

Solution for t=-5t^2+10t+20 equation:

t=-5t^2+10t+20

We move all terms to the left:

t-(-5t^2+10t+20)=0

We get rid of parentheses

5t^2-10t+t-20=0

We add all the numbers together, and all the variables

5t^2-9t-20=0

a = 5; b = -9; c = -20;
Δ = b2-4ac
Δ = -92-4·5·(-20)
Δ = 481
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{-(-9)-\sqrt{481}}{2*5}=\frac{9-\sqrt{481}}{10}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+\sqrt{481}}{2*5}=\frac{9+\sqrt{481}}{10}
$