1/4b+2/5b-5=3

Solution for 1/4b+2/5b-5=3 equation:

1/4b+2/5b-5=3

We move all terms to the left:

1/4b+2/5b-5-(3)=0


Domain of the equation: 4b!=0

b!=0/4

b!=0

b∈R



Domain of the equation: 5b!=0

b!=0/5

b!=0

b∈R


We add all the numbers together, and all the variables

1/4b+2/5b-8=0

We calculate fractions


5b/20b^2+8b/20b^2-8=0

We multiply all the terms by the denominator


5b+8b-8*20b^2=0

We add all the numbers together, and all the variables

13b-8*20b^2=0

Wy multiply elements

-160b^2+13b=0

a = -160; b = 13; c = 0;
Δ = b2-4ac
Δ = 132-4·(-160)·0
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{169}=13$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(13)-13}{2*-160}=\frac{-26}{-320}
=13/160
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+13}{2*-160}=\frac{0}{-320}
=0
$