(t-25)(t+2)=0

Solution for (t-25)(t+2)=0 equation:

(t-25)(t+2)=0

We multiply parentheses ..

(+t^2+2t-25t-50)=0

We get rid of parentheses

t^2+2t-25t-50=0

We add all the numbers together, and all the variables

t^2-23t-50=0

a = 1; b = -23; c = -50;
Δ = b2-4ac
Δ = -232-4·1·(-50)
Δ = 729
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{729}=27$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-23)-27}{2*1}=\frac{-4}{2}
=-2
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-23)+27}{2*1}=\frac{50}{2}
=25
$