1/2b+1/3b-20=-10=

Solution for 1/2b+1/3b-20=-10= equation:

1/2b+1/3b-20=-10=

We move all terms to the left:

1/2b+1/3b-20-(-10)=0


Domain of the equation: 2b!=0

b!=0/2

b!=0

b∈R



Domain of the equation: 3b!=0

b!=0/3

b!=0

b∈R


We add all the numbers together, and all the variables

1/2b+1/3b-10=0

We calculate fractions


3b/6b^2+2b/6b^2-10=0

We multiply all the terms by the denominator


3b+2b-10*6b^2=0

We add all the numbers together, and all the variables

5b-10*6b^2=0

Wy multiply elements

-60b^2+5b=0

a = -60; b = 5; c = 0;
Δ = b2-4ac
Δ = 52-4·(-60)·0
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-5}{2*-60}=\frac{-10}{-120}
=1/12
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+5}{2*-60}=\frac{0}{-120}
=0
$