(t+1)(t+1)+(t-8)(t-8)=45

Solution for (t+1)(t+1)+(t-8)(t-8)=45 equation:

(t+1)(t+1)+(t-8)(t-8)=45

We move all terms to the left:

(t+1)(t+1)+(t-8)(t-8)-(45)=0

We multiply parentheses ..

(+t^2+t+t+1)+(t-8)(t-8)-45=0

We get rid of parentheses

t^2+t+t+(t-8)(t-8)+1-45=0

We multiply parentheses ..

t^2+(+t^2-8t-8t+64)+t+t+1-45=0

We add all the numbers together, and all the variables

t^2+(+t^2-8t-8t+64)+2t-44=0

We get rid of parentheses

t^2+t^2-8t-8t+2t+64-44=0

We add all the numbers together, and all the variables

2t^2-14t+20=0

a = 2; b = -14; c = +20;
Δ = b2-4ac
Δ = -142-4·2·20
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-14)-6}{2*2}=\frac{8}{4}
=2
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-14)+6}{2*2}=\frac{20}{4}
=5
$