(t-7)(t-8)=0

Solution for (t-7)(t-8)=0 equation:

(t-7)(t-8)=0

We multiply parentheses ..

(+t^2-8t-7t+56)=0

We get rid of parentheses

t^2-8t-7t+56=0

We add all the numbers together, and all the variables

t^2-15t+56=0

a = 1; b = -15; c = +56;
Δ = b2-4ac
Δ = -152-4·1·56
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-15)-1}{2*1}=\frac{14}{2}
=7
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-15)+1}{2*1}=\frac{16}{2}
=8
$