-(4x-1)(3-x)=-11x+6

Solution for -(4x-1)(3-x)=-11x+6 equation:

-(4x-1)(3-x)=-11x+6

We move all terms to the left:

-(4x-1)(3-x)-(-11x+6)=0

We add all the numbers together, and all the variables

-(4x-1)(-1x+3)-(-11x+6)=0

We get rid of parentheses

-(4x-1)(-1x+3)+11x-6=0

We multiply parentheses ..

-(-4x^2+12x+x-3)+11x-6=0

We get rid of parentheses

4x^2-12x-x+11x+3-6=0

We add all the numbers together, and all the variables

4x^2-2x-3=0

a = 4; b = -2; c = -3;
Δ = b2-4ac
Δ = -22-4·4·(-3)
Δ = 52
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:

$\sqrt{\Delta}=\sqrt{52}=\sqrt{4*13}=\sqrt{4}*\sqrt{13}=2\sqrt{13}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-2\sqrt{13}}{2*4}=\frac{2-2\sqrt{13}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2\sqrt{13}}{2*4}=\frac{2+2\sqrt{13}}{8}
$