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

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

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

a = -5; b = -2; c = +24;
Δ = b2-4ac
Δ = -22-4·(-5)·24
Δ = 484
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{484}=22$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-22}{2*-5}=\frac{-20}{-10}
=+2
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+22}{2*-5}=\frac{24}{-10}
=-2+2/5
$