1/2b-4=2b+2

Solution for 1/2b-4=2b+2 equation:

1/2b-4=2b+2

We move all terms to the left:

1/2b-4-(2b+2)=0


Domain of the equation: 2b!=0

b!=0/2

b!=0

b∈R


We get rid of parentheses

1/2b-2b-2-4=0

We multiply all the terms by the denominator


-2b*2b-2*2b-4*2b+1=0

Wy multiply elements

-4b^2-4b-8b+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 = 160
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:

$\sqrt{\Delta}=\sqrt{160}=\sqrt{16*10}=\sqrt{16}*\sqrt{10}=4\sqrt{10}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-4\sqrt{10}}{2*-4}=\frac{12-4\sqrt{10}}{-8}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+4\sqrt{10}}{2*-4}=\frac{12+4\sqrt{10}}{-8}
$