t^2+24t-240=0

Solution for t^2+24t-240=0 equation:

t^2+24t-240=0

a = 1; b = 24; c = -240;
Δ = b2-4ac
Δ = 242-4·1·(-240)
Δ = 1536
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:

$\sqrt{\Delta}=\sqrt{1536}=\sqrt{256*6}=\sqrt{256}*\sqrt{6}=16\sqrt{6}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(24)-16\sqrt{6}}{2*1}=\frac{-24-16\sqrt{6}}{2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(24)+16\sqrt{6}}{2*1}=\frac{-24+16\sqrt{6}}{2}
$