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

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

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

We move all terms to the left:

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

We multiply parentheses

2x^2-(3(x-4))-12=0


We calculate terms in parentheses: -(3(x-4)), so:

3(x-4)

We multiply parentheses

3x-12

Back to the equation:

-(3x-12)


We get rid of parentheses

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

We add all the numbers together, and all the variables

2x^2-3x=0

a = 2; b = -3; c = 0;
Δ = b2-4ac
Δ = -32-4·2·0
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-3}{2*2}=\frac{0}{4}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+3}{2*2}=\frac{6}{4}
=1+1/2
$