(4x^2-32)=(3x^2-4x)

Solution for (4x^2-32)=(3x^2-4x) equation:

(4x^2-32)=(3x^2-4x)

We move all terms to the left:

(4x^2-32)-((3x^2-4x))=0

We get rid of parentheses

4x^2-((3x^2-4x))-32=0


We calculate terms in parentheses: -((3x^2-4x)), so:

(3x^2-4x)

We get rid of parentheses

3x^2-4x

Back to the equation:

-(3x^2-4x)


We get rid of parentheses

4x^2-3x^2+4x-32=0

We add all the numbers together, and all the variables

x^2+4x-32=0

a = 1; b = 4; c = -32;
Δ = b2-4ac
Δ = 42-4·1·(-32)
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{144}=12$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-12}{2*1}=\frac{-16}{2}
=-8
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+12}{2*1}=\frac{8}{2}
=4
$