(8-3t)(3+t)=20

Solution for (8-3t)(3+t)=20 equation:

(8-3t)(3+t)=20

We move all terms to the left:

(8-3t)(3+t)-(20)=0

We add all the numbers together, and all the variables

(-3t+8)(t+3)-20=0

We multiply parentheses ..

(-3t^2-9t+8t+24)-20=0

We get rid of parentheses

-3t^2-9t+8t+24-20=0

We add all the numbers together, and all the variables

-3t^2-1t+4=0

a = -3; b = -1; c = +4;
Δ = b2-4ac
Δ = -12-4·(-3)·4
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-7}{2*-3}=\frac{-6}{-6}
=1
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+7}{2*-3}=\frac{8}{-6}
=-1+1/3
$