(2x-3)(x+3)=(x+6)(3x+1)

Solution for (2x-3)(x+3)=(x+6)(3x+1) equation:

(2x-3)(x+3)=(x+6)(3x+1)

We move all terms to the left:

(2x-3)(x+3)-((x+6)(3x+1))=0

We multiply parentheses ..

(+2x^2+6x-3x-9)-((x+6)(3x+1))=0


We calculate terms in parentheses: -((x+6)(3x+1)), so:

(x+6)(3x+1)

We multiply parentheses ..

(+3x^2+x+18x+6)

We get rid of parentheses

3x^2+x+18x+6

We add all the numbers together, and all the variables

3x^2+19x+6

Back to the equation:

-(3x^2+19x+6)


We get rid of parentheses

2x^2-3x^2+6x-3x-19x-9-6=0

We add all the numbers together, and all the variables

-1x^2-16x-15=0

a = -1; b = -16; c = -15;
Δ = b2-4ac
Δ = -162-4·(-1)·(-15)
Δ = 196
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{196}=14$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-14}{2*-1}=\frac{2}{-2}
=-1
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+14}{2*-1}=\frac{30}{-2}
=-15
$