-17+20=-18+3b-5/2b

Solution for -17+20=-18+3b-5/2b equation:

-17+20=-18+3b-5/2b

We move all terms to the left:

-17+20-(-18+3b-5/2b)=0


Domain of the equation: 2b)!=0

b!=0/1

b!=0

b∈R


We add all the numbers together, and all the variables

-(3b-5/2b-18)-17+20=0

We add all the numbers together, and all the variables

-(3b-5/2b-18)+3=0

We get rid of parentheses

-3b+5/2b+18+3=0

We multiply all the terms by the denominator


-3b*2b+18*2b+3*2b+5=0

Wy multiply elements

-6b^2+36b+6b+5=0

We add all the numbers together, and all the variables

-6b^2+42b+5=0

a = -6; b = 42; c = +5;
Δ = b2-4ac
Δ = 422-4·(-6)·5
Δ = 1884
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:

$\sqrt{\Delta}=\sqrt{1884}=\sqrt{4*471}=\sqrt{4}*\sqrt{471}=2\sqrt{471}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(42)-2\sqrt{471}}{2*-6}=\frac{-42-2\sqrt{471}}{-12}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(42)+2\sqrt{471}}{2*-6}=\frac{-42+2\sqrt{471}}{-12}
$