(4x-8)=(4x^2-16)

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

(4x-8)=(4x^2-16)

We move all terms to the left:

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

We get rid of parentheses

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


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

(4x^2-16)

We get rid of parentheses

4x^2-16

Back to the equation:

-(4x^2-16)


We get rid of parentheses

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

We add all the numbers together, and all the variables

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

a = -4; b = 4; c = +8;
Δ = b2-4ac
Δ = 42-4·(-4)·8
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*-4}=\frac{-16}{-8}
=+2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+12}{2*-4}=\frac{8}{-8}
=-1
$