t^2+10t-135=0

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

t^2+10t-135=0

a = 1; b = 10; c = -135;
Δ = b2-4ac
Δ = 102-4·1·(-135)
Δ = 640
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:

$\sqrt{\Delta}=\sqrt{640}=\sqrt{64*10}=\sqrt{64}*\sqrt{10}=8\sqrt{10}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-8\sqrt{10}}{2*1}=\frac{-10-8\sqrt{10}}{2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+8\sqrt{10}}{2*1}=\frac{-10+8\sqrt{10}}{2}
$