-5t^2+40t-35=20

Solution for -5t^2+40t-35=20 equation:

-5t^2+40t-35=20

We move all terms to the left:

-5t^2+40t-35-(20)=0

We add all the numbers together, and all the variables

-5t^2+40t-55=0

a = -5; b = 40; c = -55;
Δ = b2-4ac
Δ = 402-4·(-5)·(-55)
Δ = 500
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:

$\sqrt{\Delta}=\sqrt{500}=\sqrt{100*5}=\sqrt{100}*\sqrt{5}=10\sqrt{5}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-10\sqrt{5}}{2*-5}=\frac{-40-10\sqrt{5}}{-10}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+10\sqrt{5}}{2*-5}=\frac{-40+10\sqrt{5}}{-10}
$