-15t^2+135t+10=0

Solution for -15t^2+135t+10=0 equation:

-15t^2+135t+10=0

a = -15; b = 135; c = +10;
Δ = b2-4ac
Δ = 1352-4·(-15)·10
Δ = 18825
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:

$\sqrt{\Delta}=\sqrt{18825}=\sqrt{25*753}=\sqrt{25}*\sqrt{753}=5\sqrt{753}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(135)-5\sqrt{753}}{2*-15}=\frac{-135-5\sqrt{753}}{-30}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(135)+5\sqrt{753}}{2*-15}=\frac{-135+5\sqrt{753}}{-30}
$