(a-5)(a+3)=105

Solution for (a-5)(a+3)=105 equation:

(a-5)(a+3)=105

We move all terms to the left:

(a-5)(a+3)-(105)=0

We multiply parentheses ..

(+a^2+3a-5a-15)-105=0

We get rid of parentheses

a^2+3a-5a-15-105=0

We add all the numbers together, and all the variables

a^2-2a-120=0

a = 1; b = -2; c = -120;
Δ = b2-4ac
Δ = -22-4·1·(-120)
Δ = 484
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{484}=22$
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-22}{2*1}=\frac{-20}{2}
=-10
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+22}{2*1}=\frac{24}{2}
=12
$