(t-5)(t+7)=0

Solution for (t-5)(t+7)=0 equation:

(t-5)(t+7)=0

We multiply parentheses ..

(+t^2+7t-5t-35)=0

We get rid of parentheses

t^2+7t-5t-35=0

We add all the numbers together, and all the variables

t^2+2t-35=0

a = 1; b = 2; c = -35;
Δ = b2-4ac
Δ = 22-4·1·(-35)
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{144}=12$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-12}{2*1}=\frac{-14}{2}
=-7
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+12}{2*1}=\frac{10}{2}
=5
$