5t+5t^2=40

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

5t+5t^2=40

We move all terms to the left:

5t+5t^2-(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}
$