12x^2-3x+4x=6

Solution for 12x^2-3x+4x=6 equation:

12x^2-3x+4x=6

We move all terms to the left:

12x^2-3x+4x-(6)=0

We add all the numbers together, and all the variables

12x^2+x-6=0

a = 12; b = 1; c = -6;
Δ = b2-4ac
Δ = 12-4·12·(-6)
Δ = 289
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{289}=17$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-17}{2*12}=\frac{-18}{24}
=-3/4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+17}{2*12}=\frac{16}{24}
=2/3
$