(t+8)(t+3)=4

Solution for (t+8)(t+3)=4 equation:

(t+8)(t+3)=4

We move all terms to the left:

(t+8)(t+3)-(4)=0

We multiply parentheses ..

(+t^2+3t+8t+24)-4=0

We get rid of parentheses

t^2+3t+8t+24-4=0

We add all the numbers together, and all the variables

t^2+11t+20=0

a = 1; b = 11; c = +20;
Δ = b2-4ac
Δ = 112-4·1·20
Δ = 41
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(11)-\sqrt{41}}{2*1}=\frac{-11-\sqrt{41}}{2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+\sqrt{41}}{2*1}=\frac{-11+\sqrt{41}}{2}
$