5t^2+24t-96=0

Solution for 5t^2+24t-96=0 equation:

5t^2+24t-96=0

a = 5; b = 24; c = -96;
Δ = b2-4ac
Δ = 242-4·5·(-96)
Δ = 2496
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:

$\sqrt{\Delta}=\sqrt{2496}=\sqrt{64*39}=\sqrt{64}*\sqrt{39}=8\sqrt{39}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(24)-8\sqrt{39}}{2*5}=\frac{-24-8\sqrt{39}}{10}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(24)+8\sqrt{39}}{2*5}=\frac{-24+8\sqrt{39}}{10}
$