t2+20t-125=0

Solution for t2+20t-125=0 equation:

t2+20t-125=0

We add all the numbers together, and all the variables

t^2+20t-125=0

a = 1; b = 20; c = -125;
Δ = b2-4ac
Δ = 202-4·1·(-125)
Δ = 900
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{900}=30$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-30}{2*1}=\frac{-50}{2}
=-25
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+30}{2*1}=\frac{10}{2}
=5
$