(1/2)(4b+2)=5b-4-3b

Solution for (1/2)(4b+2)=5b-4-3b equation:

(1/2)(4b+2)=5b-4-3b

We move all terms to the left:

(1/2)(4b+2)-(5b-4-3b)=0


Domain of the equation: 2)(4b+2)!=0

b∈R


We add all the numbers together, and all the variables

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

We get rid of parentheses

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

We multiply parentheses ..

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

We multiply all the terms by the denominator


(+4b^2+1-2b*2*2)+4*2*2)=0

We add all the numbers together, and all the variables

(+4b^2+1-2b*2*2)=0

We get rid of parentheses

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

Wy multiply elements

4b^2-8b*2+1=0

Wy multiply elements

4b^2-16b+1=0

a = 4; b = -16; c = +1;
Δ = b2-4ac
Δ = -162-4·4·1
Δ = 240
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:

$\sqrt{\Delta}=\sqrt{240}=\sqrt{16*15}=\sqrt{16}*\sqrt{15}=4\sqrt{15}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-4\sqrt{15}}{2*4}=\frac{16-4\sqrt{15}}{8}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+4\sqrt{15}}{2*4}=\frac{16+4\sqrt{15}}{8}
$