0=5t^2+5t-40

Solution for 0=5t^2+5t-40 equation:

0=5t^2+5t-40

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We get rid of parentheses

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

a = -5; b = -5; c = +40;
Δ = b2-4ac
Δ = -52-4·(-5)·40
Δ = 825
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:

$\sqrt{\Delta}=\sqrt{825}=\sqrt{25*33}=\sqrt{25}*\sqrt{33}=5\sqrt{33}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-5\sqrt{33}}{2*-5}=\frac{5-5\sqrt{33}}{-10}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+5\sqrt{33}}{2*-5}=\frac{5+5\sqrt{33}}{-10}
$