X^2+4x=(2x+3)(x-2)

Solution for X^2+4x=(2x+3)(x-2) equation:

X^2+4X=(2X+3)(X-2)

We move all terms to the left:

X^2+4X-((2X+3)(X-2))=0

We multiply parentheses ..

X^2-((+2X^2-4X+3X-6))+4X=0


We calculate terms in parentheses: -((+2X^2-4X+3X-6)), so:

(+2X^2-4X+3X-6)

We get rid of parentheses

2X^2-4X+3X-6

We add all the numbers together, and all the variables

2X^2-1X-6

Back to the equation:

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


We add all the numbers together, and all the variables

X^2+4X-(2X^2-1X-6)=0

We get rid of parentheses

X^2-2X^2+4X+1X+6=0

We add all the numbers together, and all the variables

-1X^2+5X+6=0

a = -1; b = 5; c = +6;
Δ = b2-4ac
Δ = 52-4·(-1)·6
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$
$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-7}{2*-1}=\frac{-12}{-2}
=+6
$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+7}{2*-1}=\frac{2}{-2}
=-1
$