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

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

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

We move all terms to the left:

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

We multiply parentheses ..

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


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

6(x-2)

We multiply parentheses

6x-12

Back to the equation:

-(6x-12)


We get rid of parentheses

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

We add all the numbers together, and all the variables

2x^2-1x=0

a = 2; b = -1; c = 0;
Δ = b2-4ac
Δ = -12-4·2·0
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-1}{2*2}=\frac{0}{4}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+1}{2*2}=\frac{2}{4}
=1/2
$