5x(x-4)-3x(4x+1)=-x^2+3x-18

Solution for 5x(x-4)-3x(4x+1)=-x^2+3x-18 equation:

5x(x-4)-3x(4x+1)=-x^2+3x-18

We move all terms to the left:

5x(x-4)-3x(4x+1)-(-x^2+3x-18)=0

We multiply parentheses

-(-x^2+3x-18)+5x^2-12x^2-20x-3x=0

We get rid of parentheses

x^2+5x^2-12x^2-3x-20x-3x+18=0

We add all the numbers together, and all the variables

-6x^2-26x+18=0

a = -6; b = -26; c = +18;
Δ = b2-4ac
Δ = -262-4·(-6)·18
Δ = 1108
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:

$\sqrt{\Delta}=\sqrt{1108}=\sqrt{4*277}=\sqrt{4}*\sqrt{277}=2\sqrt{277}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-26)-2\sqrt{277}}{2*-6}=\frac{26-2\sqrt{277}}{-12}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-26)+2\sqrt{277}}{2*-6}=\frac{26+2\sqrt{277}}{-12}
$