-2b+8=1/2b+2

Solution for -2b+8=1/2b+2 equation:

-2b+8=1/2b+2

We move all terms to the left:

-2b+8-(1/2b+2)=0


Domain of the equation: 2b+2)!=0

b∈R


We get rid of parentheses

-2b-1/2b-2+8=0

We multiply all the terms by the denominator


-2b*2b-2*2b+8*2b-1=0

Wy multiply elements

-4b^2-4b+16b-1=0

We add all the numbers together, and all the variables

-4b^2+12b-1=0

a = -4; b = 12; c = -1;
Δ = b2-4ac
Δ = 122-4·(-4)·(-1)
Δ = 128
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{128}=\sqrt{64*2}=\sqrt{64}*\sqrt{2}=8\sqrt{2}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-8\sqrt{2}}{2*-4}=\frac{-12-8\sqrt{2}}{-8}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+8\sqrt{2}}{2*-4}=\frac{-12+8\sqrt{2}}{-8}
$