0=(8-2x)(11-2x)

Solution for 0=(8-2x)(11-2x) equation:

0=(8-2x)(11-2x)

We move all terms to the left:

0-((8-2x)(11-2x))=0

We add all the numbers together, and all the variables

-((-2x+8)(-2x+11))+0=0

We add all the numbers together, and all the variables

-((-2x+8)(-2x+11))=0

We multiply parentheses ..

-((+4x^2-22x-16x+88))=0


We calculate terms in parentheses: -((+4x^2-22x-16x+88)), so:

(+4x^2-22x-16x+88)

We get rid of parentheses

4x^2-22x-16x+88

We add all the numbers together, and all the variables

4x^2-38x+88

Back to the equation:

-(4x^2-38x+88)


We get rid of parentheses

-4x^2+38x-88=0

a = -4; b = 38; c = -88;
Δ = b2-4ac
Δ = 382-4·(-4)·(-88)
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(38)-6}{2*-4}=\frac{-44}{-8}
=5+1/2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(38)+6}{2*-4}=\frac{-32}{-8}
=+4
$