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

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

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

We move all terms to the left:

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

We get rid of parentheses

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

We multiply parentheses ..

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

We add all the numbers together, and all the variables

(+t^2+3t-8t-24)-1t+8=0

We get rid of parentheses

t^2+3t-8t-1t-24+8=0

We add all the numbers together, and all the variables

t^2-6t-16=0

a = 1; b = -6; c = -16;
Δ = b2-4ac
Δ = -62-4·1·(-16)
Δ = 100
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{100}=10$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-10}{2*1}=\frac{-4}{2}
=-2
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+10}{2*1}=\frac{16}{2}
=8
$