(2x+1)(x-4)-x^2+8x-16=0

Solution for (2x+1)(x-4)-x^2+8x-16=0 equation:

(2x+1)(x-4)-x^2+8x-16=0

We add all the numbers together, and all the variables

-1x^2+8x+(2x+1)(x-4)-16=0

We multiply parentheses ..

-1x^2+(+2x^2-8x+x-4)+8x-16=0

We get rid of parentheses

-1x^2+2x^2-8x+x+8x-4-16=0

We add all the numbers together, and all the variables

x^2+x-20=0

a = 1; b = 1; c = -20;
Δ = b2-4ac
Δ = 12-4·1·(-20)
Δ = 81
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{81}=9$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-9}{2*1}=\frac{-10}{2}
=-5
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+9}{2*1}=\frac{8}{2}
=4
$