10-4/5b=3/4b-12

Solution for 10-4/5b=3/4b-12 equation:

10-4/5b=3/4b-12

We move all terms to the left:

10-4/5b-(3/4b-12)=0


Domain of the equation: 5b!=0

b!=0/5

b!=0

b∈R



Domain of the equation: 4b-12)!=0

b∈R


We get rid of parentheses

-4/5b-3/4b+12+10=0

We calculate fractions


(-16b)/20b^2+(-15b)/20b^2+12+10=0

We add all the numbers together, and all the variables

(-16b)/20b^2+(-15b)/20b^2+22=0

We multiply all the terms by the denominator


(-16b)+(-15b)+22*20b^2=0

Wy multiply elements

440b^2+(-16b)+(-15b)=0

We get rid of parentheses

440b^2-16b-15b=0

We add all the numbers together, and all the variables

440b^2-31b=0

a = 440; b = -31; c = 0;
Δ = b2-4ac
Δ = -312-4·440·0
Δ = 961
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{961}=31$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-31)-31}{2*440}=\frac{0}{880}
=0
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-31)+31}{2*440}=\frac{62}{880}
=31/440
$