-t^2-13t+40=0

Solution for -t^2-13t+40=0 equation:

-t^2-13t+40=0

We add all the numbers together, and all the variables

-1t^2-13t+40=0

a = -1; b = -13; c = +40;
Δ = b2-4ac
Δ = -132-4·(-1)·40
Δ = 329
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-13)-\sqrt{329}}{2*-1}=\frac{13-\sqrt{329}}{-2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-13)+\sqrt{329}}{2*-1}=\frac{13+\sqrt{329}}{-2}
$