-5t^2+40t=-20

Solution for -5t^2+40t=-20 equation:

-5t^2+40t=-20

We move all terms to the left:

-5t^2+40t-(-20)=0

We add all the numbers together, and all the variables

-5t^2+40t+20=0

a = -5; b = 40; c = +20;
Δ = b2-4ac
Δ = 402-4·(-5)·20
Δ = 2000
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{2000}=\sqrt{400*5}=\sqrt{400}*\sqrt{5}=20\sqrt{5}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-20\sqrt{5}}{2*-5}=\frac{-40-20\sqrt{5}}{-10}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+20\sqrt{5}}{2*-5}=\frac{-40+20\sqrt{5}}{-10}
$