(a+2)(a+3)=15

Solution for (a+2)(a+3)=15 equation:

(a+2)(a+3)=15

We move all terms to the left:

(a+2)(a+3)-(15)=0

We multiply parentheses ..

(+a^2+3a+2a+6)-15=0

We get rid of parentheses

a^2+3a+2a+6-15=0

We add all the numbers together, and all the variables

a^2+5a-9=0

a = 1; b = 5; c = -9;
Δ = b2-4ac
Δ = 52-4·1·(-9)
Δ = 61
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-\sqrt{61}}{2*1}=\frac{-5-\sqrt{61}}{2}
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+\sqrt{61}}{2*1}=\frac{-5+\sqrt{61}}{2}
$