(t+5)(t-12)=0

Solution for (t+5)(t-12)=0 equation:

(t+5)(t-12)=0

We multiply parentheses ..

(+t^2-12t+5t-60)=0

We get rid of parentheses

t^2-12t+5t-60=0

We add all the numbers together, and all the variables

t^2-7t-60=0

a = 1; b = -7; c = -60;
Δ = b2-4ac
Δ = -72-4·1·(-60)
Δ = 289
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{289}=17$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-7)-17}{2*1}=\frac{-10}{2}
=-5
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-7)+17}{2*1}=\frac{24}{2}
=12
$