(X+2)(x+3)=15

Solution for (X+2)(x+3)=15 equation:

(X+2)(X+3)=15

We move all terms to the left:

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

We multiply parentheses ..

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

X^2+5X-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:
$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

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