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

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



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

We move all terms to the left:

2x-3-(3(2+x)(-3+x))=0

We add all the numbers together, and all the variables

2x-(3(x+2)(x-3))-3=0

We multiply parentheses ..

-(3(+x^2-3x+2x-6))+2x-3=0



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

3(+x^2-3x+2x-6)

We multiply parentheses

3x^2-9x+6x-18

We add all the numbers together, and all the variables

3x^2-3x-18

Back to the equation:

-(3x^2-3x-18)



We add all the numbers together, and all the variables

2x-(3x^2-3x-18)-3=0

We get rid of parentheses

-3x^2+2x+3x+18-3=0

We add all the numbers together, and all the variables

-3x^2+5x+15=0

a = -3; b = 5; c = +15;
Δ = b2-4ac
Δ = 52-4·(-3)·15
Δ = 205
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{205}}{2*-3}=\frac{-5-\sqrt{205}}{-6}
$


$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+\sqrt{205}}{2*-3}=\frac{-5+\sqrt{205}}{-6}
$

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