-5t^2-3t+26=0

Solution for -5t^2-3t+26=0 equation:

-5t^2-3t+26=0

a = -5; b = -3; c = +26;
Δ = b2-4ac
Δ = -32-4·(-5)·26
Δ = 529
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{529}=23$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-23}{2*-5}=\frac{-20}{-10}
=+2
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+23}{2*-5}=\frac{26}{-10}
=-2+3/5
$