5t^2-26t-40=0

Solution for 5t^2-26t-40=0 equation:

5t^2-26t-40=0

a = 5; b = -26; c = -40;
Δ = b2-4ac
Δ = -262-4·5·(-40)
Δ = 1476
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:

$\sqrt{\Delta}=\sqrt{1476}=\sqrt{36*41}=\sqrt{36}*\sqrt{41}=6\sqrt{41}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-26)-6\sqrt{41}}{2*5}=\frac{26-6\sqrt{41}}{10}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-26)+6\sqrt{41}}{2*5}=\frac{26+6\sqrt{41}}{10}
$