11/2b+2=3/5b+10

Solution for 11/2b+2=3/5b+10 equation:

11/2b+2=3/5b+10

We move all terms to the left:

11/2b+2-(3/5b+10)=0


Domain of the equation: 2b!=0

b!=0/2

b!=0

b∈R



Domain of the equation: 5b+10)!=0

b∈R


We get rid of parentheses

11/2b-3/5b-10+2=0

We calculate fractions


55b/10b^2+(-6b)/10b^2-10+2=0

We add all the numbers together, and all the variables

55b/10b^2+(-6b)/10b^2-8=0

We multiply all the terms by the denominator


55b+(-6b)-8*10b^2=0

Wy multiply elements

-80b^2+55b+(-6b)=0

We get rid of parentheses

-80b^2+55b-6b=0

We add all the numbers together, and all the variables

-80b^2+49b=0

a = -80; b = 49; c = 0;
Δ = b2-4ac
Δ = 492-4·(-80)·0
Δ = 2401
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2401}=49$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(49)-49}{2*-80}=\frac{-98}{-160}
=49/80
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(49)+49}{2*-80}=\frac{0}{-160}
=0
$