4*(b^+3)=b*(4b+2)

Solution for 4*(b^+3)=b*(4b+2) equation:

4(b^+3)=b(4b+2)

We move all terms to the left:

4(b^+3)-(b(4b+2))=0

We multiply parentheses

4b-(b(4b+2))+12=0


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

b(4b+2)

We multiply parentheses

4b^2+2b

Back to the equation:

-(4b^2+2b)


We get rid of parentheses

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

We add all the numbers together, and all the variables

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

a = -4; b = 2; c = +12;
Δ = b2-4ac
Δ = 22-4·(-4)·12
Δ = 196
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{196}=14$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-14}{2*-4}=\frac{-16}{-8}
=+2
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+14}{2*-4}=\frac{12}{-8}
=-1+1/2
$