(2x^2+8)+(x-1)=(8x+2)

Solution for (2x^2+8)+(x-1)=(8x+2) equation:

(2x^2+8)+(x-1)=(8x+2)

We move all terms to the left:

(2x^2+8)+(x-1)-((8x+2))=0

We get rid of parentheses

2x^2+x-((8x+2))+8-1=0


We calculate terms in parentheses: -((8x+2)), so:

(8x+2)

We get rid of parentheses

8x+2

Back to the equation:

-(8x+2)


We add all the numbers together, and all the variables

2x^2+x-(8x+2)+7=0

We get rid of parentheses

2x^2+x-8x-2+7=0

We add all the numbers together, and all the variables

2x^2-7x+5=0

a = 2; b = -7; c = +5;
Δ = b2-4ac
Δ = -72-4·2·5
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-7)-3}{2*2}=\frac{4}{4}
=1
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-7)+3}{2*2}=\frac{10}{4}
=2+1/2
$