3x-4x2+6=x-4

Solution for 3x-4x2+6=x-4 equation:

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

We move all terms to the left:

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

We get rid of parentheses

-4x^2+3x-x+4+6=0

We add all the numbers together, and all the variables

-4x^2+2x+10=0

a = -4; b = 2; c = +10;
Δ = b2-4ac
Δ = 22-4·(-4)·10
Δ = 164
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{164}=\sqrt{4*41}=\sqrt{4}*\sqrt{41}=2\sqrt{41}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{41}}{2*-4}=\frac{-2-2\sqrt{41}}{-8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{41}}{2*-4}=\frac{-2+2\sqrt{41}}{-8}
$