x^2=(7x-32)(3x-16)

Solution for x^2=(7x-32)(3x-16) equation:

x^2=(7x-32)(3x-16)

We move all terms to the left:

x^2-((7x-32)(3x-16))=0

We multiply parentheses ..

x^2-((+21x^2-112x-96x+512))=0


We calculate terms in parentheses: -((+21x^2-112x-96x+512)), so:

(+21x^2-112x-96x+512)

We get rid of parentheses

21x^2-112x-96x+512

We add all the numbers together, and all the variables

21x^2-208x+512

Back to the equation:

-(21x^2-208x+512)


We get rid of parentheses

x^2-21x^2+208x-512=0

We add all the numbers together, and all the variables

-20x^2+208x-512=0

a = -20; b = 208; c = -512;
Δ = b2-4ac
Δ = 2082-4·(-20)·(-512)
Δ = 2304
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2304}=48$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(208)-48}{2*-20}=\frac{-256}{-40}
=6+2/5
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(208)+48}{2*-20}=\frac{-160}{-40}
=+4
$