(X+2)(X+3)=(X+2)(3x-1)

Solution for (X+2)(X+3)=(X+2)(3x-1) equation:

(X+2)(X+3)=(X+2)(3X-1)

We move all terms to the left:

(X+2)(X+3)-((X+2)(3X-1))=0

We multiply parentheses ..

(+X^2+3X+2X+6)-((X+2)(3X-1))=0


We calculate terms in parentheses: -((X+2)(3X-1)), so:

(X+2)(3X-1)

We multiply parentheses ..

(+3X^2-1X+6X-2)

We get rid of parentheses

3X^2-1X+6X-2

We add all the numbers together, and all the variables

3X^2+5X-2

Back to the equation:

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


We get rid of parentheses

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

We add all the numbers together, and all the variables

-2X^2+8=0

a = -2; b = 0; c = +8;
Δ = b2-4ac
Δ = 02-4·(-2)·8
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8}{2*-2}=\frac{-8}{-4}
=+2
$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8}{2*-2}=\frac{8}{-4}
=-2
$