(t-2)(t+7)=0

Solution for (t-2)(t+7)=0 equation:

(t-2)(t+7)=0

We multiply parentheses ..

(+t^2+7t-2t-14)=0

We get rid of parentheses

t^2+7t-2t-14=0

We add all the numbers together, and all the variables

t^2+5t-14=0

a = 1; b = 5; c = -14;
Δ = b2-4ac
Δ = 52-4·1·(-14)
Δ = 81
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{81}=9$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-9}{2*1}=\frac{-14}{2}
=-7
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+9}{2*1}=\frac{4}{2}
=2
$