-5t^2+56t+24=0

Solution for -5t^2+56t+24=0 equation:

-5t^2+56t+24=0

a = -5; b = 56; c = +24;
Δ = b2-4ac
Δ = 562-4·(-5)·24
Δ = 3616
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:

$\sqrt{\Delta}=\sqrt{3616}=\sqrt{16*226}=\sqrt{16}*\sqrt{226}=4\sqrt{226}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(56)-4\sqrt{226}}{2*-5}=\frac{-56-4\sqrt{226}}{-10}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(56)+4\sqrt{226}}{2*-5}=\frac{-56+4\sqrt{226}}{-10}
$