5t^2+20t-40=0

Solution for 5t^2+20t-40=0 equation:

5t^2+20t-40=0

a = 5; b = 20; c = -40;
Δ = b2-4ac
Δ = 202-4·5·(-40)
Δ = 1200
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:

$\sqrt{\Delta}=\sqrt{1200}=\sqrt{400*3}=\sqrt{400}*\sqrt{3}=20\sqrt{3}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-20\sqrt{3}}{2*5}=\frac{-20-20\sqrt{3}}{10}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+20\sqrt{3}}{2*5}=\frac{-20+20\sqrt{3}}{10}
$