11b+2=3/5b+10

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

11b+2=3/5b+10

We move all terms to the left:

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


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

b∈R


We get rid of parentheses

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

We multiply all the terms by the denominator


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

Wy multiply elements

55b^2-50b+10b-3=0

We add all the numbers together, and all the variables

55b^2-40b-3=0

a = 55; b = -40; c = -3;
Δ = b2-4ac
Δ = -402-4·55·(-3)
Δ = 2260
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:

$\sqrt{\Delta}=\sqrt{2260}=\sqrt{4*565}=\sqrt{4}*\sqrt{565}=2\sqrt{565}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-40)-2\sqrt{565}}{2*55}=\frac{40-2\sqrt{565}}{110}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-40)+2\sqrt{565}}{2*55}=\frac{40+2\sqrt{565}}{110}
$