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

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

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

We move all terms to the left:

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

We multiply parentheses ..

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

We add all the numbers together, and all the variables

3x^2+5x-23=0

a = 3; b = 5; c = -23;
Δ = b2-4ac
Δ = 52-4·3·(-23)
Δ = 301
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{301}}{2*3}=\frac{-5-\sqrt{301}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+\sqrt{301}}{2*3}=\frac{-5+\sqrt{301}}{6}
$