-5t^2-26t+24=0

Solution for -5t^2-26t+24=0 equation:

-5t^2-26t+24=0

a = -5; b = -26; c = +24;
Δ = b2-4ac
Δ = -262-4·(-5)·24
Δ = 1156
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1156}=34$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-26)-34}{2*-5}=\frac{-8}{-10}
=4/5
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-26)+34}{2*-5}=\frac{60}{-10}
=-6
$