(8-3x)(5x+2)=4x(11-4x)

Solution for (8-3x)(5x+2)=4x(11-4x) equation:

(8-3x)(5x+2)=4x(11-4x)

We move all terms to the left:

(8-3x)(5x+2)-(4x(11-4x))=0

We add all the numbers together, and all the variables

(-3x+8)(5x+2)-(4x(-4x+11))=0

We multiply parentheses ..

(-15x^2-6x+40x+16)-(4x(-4x+11))=0


We calculate terms in parentheses: -(4x(-4x+11)), so:

4x(-4x+11)

We multiply parentheses

-16x^2+44x

Back to the equation:

-(-16x^2+44x)


We get rid of parentheses

-15x^2+16x^2-6x+40x-44x+16=0

We add all the numbers together, and all the variables

x^2-10x+16=0

a = 1; b = -10; c = +16;
Δ = b2-4ac
Δ = -102-4·1·16
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{-(-10)-6}{2*1}=\frac{4}{2}
=2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+6}{2*1}=\frac{16}{2}
=8
$