(3x-4)(2x+5)-15x=x(3x+2)-12

Solution for (3x-4)(2x+5)-15x=x(3x+2)-12 equation:

(3x-4)(2x+5)-15x=x(3x+2)-12

We move all terms to the left:

(3x-4)(2x+5)-15x-(x(3x+2)-12)=0

We add all the numbers together, and all the variables

-15x+(3x-4)(2x+5)-(x(3x+2)-12)=0

We multiply parentheses ..

(+6x^2+15x-8x-20)-15x-(x(3x+2)-12)=0


We calculate terms in parentheses: -(x(3x+2)-12), so:

x(3x+2)-12

We multiply parentheses

3x^2+2x-12

Back to the equation:

-(3x^2+2x-12)


We get rid of parentheses

6x^2-3x^2+15x-8x-15x-2x-20+12=0

We add all the numbers together, and all the variables

3x^2-10x-8=0

a = 3; b = -10; c = -8;
Δ = b2-4ac
Δ = -102-4·3·(-8)
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{-(-10)-14}{2*3}=\frac{-4}{6}
=-2/3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+14}{2*3}=\frac{24}{6}
=4
$