(t+3)(t-8)=0

Solution for (t+3)(t-8)=0 equation:

(t+3)(t-8)=0

We multiply parentheses ..

(+t^2-8t+3t-24)=0

We get rid of parentheses

t^2-8t+3t-24=0

We add all the numbers together, and all the variables

t^2-5t-24=0

a = 1; b = -5; c = -24;
Δ = b2-4ac
Δ = -52-4·1·(-24)
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-11}{2*1}=\frac{-6}{2}
=-3
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+11}{2*1}=\frac{16}{2}
=8
$