(x+4)(2x-3)-(5x-6)(x-3)=10

Solution for (x+4)(2x-3)-(5x-6)(x-3)=10 equation:

(x+4)(2x-3)-(5x-6)(x-3)=10

We move all terms to the left:

(x+4)(2x-3)-(5x-6)(x-3)-(10)=0

We multiply parentheses ..

(+2x^2-3x+8x-12)-(5x-6)(x-3)-10=0

We get rid of parentheses

2x^2-3x+8x-(5x-6)(x-3)-12-10=0

We multiply parentheses ..

2x^2-(+5x^2-15x-6x+18)-3x+8x-12-10=0

We add all the numbers together, and all the variables

2x^2-(+5x^2-15x-6x+18)+5x-22=0

We get rid of parentheses

2x^2-5x^2+15x+6x+5x-18-22=0

We add all the numbers together, and all the variables

-3x^2+26x-40=0

a = -3; b = 26; c = -40;
Δ = b2-4ac
Δ = 262-4·(-3)·(-40)
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{-(26)-14}{2*-3}=\frac{-40}{-6}
=6+2/3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(26)+14}{2*-3}=\frac{-12}{-6}
=+2
$