b-5=3/2b+5

Solution for b-5=3/2b+5 equation:

b-5=3/2b+5

We move all terms to the left:

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


Domain of the equation: 2b+5)!=0

b∈R


We get rid of parentheses

b-3/2b-5-5=0

We multiply all the terms by the denominator


b*2b-5*2b-5*2b-3=0

Wy multiply elements

2b^2-10b-10b-3=0

We add all the numbers together, and all the variables

2b^2-20b-3=0

a = 2; b = -20; c = -3;
Δ = b2-4ac
Δ = -202-4·2·(-3)
Δ = 424
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:

$\sqrt{\Delta}=\sqrt{424}=\sqrt{4*106}=\sqrt{4}*\sqrt{106}=2\sqrt{106}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-2\sqrt{106}}{2*2}=\frac{20-2\sqrt{106}}{4}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+2\sqrt{106}}{2*2}=\frac{20+2\sqrt{106}}{4}
$