5t^2+10t-135=0

Solution for 5t^2+10t-135=0 equation:

5t^2+10t-135=0

a = 5; b = 10; c = -135;
Δ = b2-4ac
Δ = 102-4·5·(-135)
Δ = 2800
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:

$\sqrt{\Delta}=\sqrt{2800}=\sqrt{400*7}=\sqrt{400}*\sqrt{7}=20\sqrt{7}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-20\sqrt{7}}{2*5}=\frac{-10-20\sqrt{7}}{10}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+20\sqrt{7}}{2*5}=\frac{-10+20\sqrt{7}}{10}
$