-5t^2+24t+7=0

Solution for -5t^2+24t+7=0 equation:

-5t^2+24t+7=0

a = -5; b = 24; c = +7;
Δ = b2-4ac
Δ = 242-4·(-5)·7
Δ = 716
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:

$\sqrt{\Delta}=\sqrt{716}=\sqrt{4*179}=\sqrt{4}*\sqrt{179}=2\sqrt{179}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(24)-2\sqrt{179}}{2*-5}=\frac{-24-2\sqrt{179}}{-10}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(24)+2\sqrt{179}}{2*-5}=\frac{-24+2\sqrt{179}}{-10}
$