2x^2+8=2(x+4)

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

2x^2+8=2(x+4)

We move all terms to the left:

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


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

2(x+4)

We multiply parentheses

2x+8

Back to the equation:

-(2x+8)


We get rid of parentheses

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

We add all the numbers together, and all the variables

2x^2-2x=0

a = 2; b = -2; c = 0;
Δ = b2-4ac
Δ = -22-4·2·0
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-2}{2*2}=\frac{0}{4}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2}{2*2}=\frac{4}{4}
=1
$