(X+3)(x-5)=2(x+3)

Solution for (X+3)(x-5)=2(x+3) equation:

(X+3)(X-5)=2(X+3)

We move all terms to the left:

(X+3)(X-5)-(2(X+3))=0

We multiply parentheses ..

(+X^2-5X+3X-15)-(2(X+3))=0


We calculate terms in parentheses: -(2(X+3)), so:

2(X+3)

We multiply parentheses

2X+6

Back to the equation:

-(2X+6)


We get rid of parentheses

X^2-5X+3X-2X-15-6=0

We add all the numbers together, and all the variables

X^2-4X-21=0

a = 1; b = -4; c = -21;
Δ = b2-4ac
Δ = -42-4·1·(-21)
Δ = 100
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{100}=10$
$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-10}{2*1}=\frac{-6}{2}
=-3
$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+10}{2*1}=\frac{14}{2}
=7
$