3/5b-2+7b-12-2/5b=14

Solution for 3/5b-2+7b-12-2/5b=14 equation:

3/5b-2+7b-12-2/5b=14

We move all terms to the left:

3/5b-2+7b-12-2/5b-(14)=0


Domain of the equation: 5b!=0

b!=0/5

b!=0

b∈R


We add all the numbers together, and all the variables

7b+3/5b-2/5b-28=0

We multiply all the terms by the denominator


7b*5b-28*5b+3-2=0

We add all the numbers together, and all the variables

7b*5b-28*5b+1=0

Wy multiply elements

35b^2-140b+1=0

a = 35; b = -140; c = +1;
Δ = b2-4ac
Δ = -1402-4·35·1
Δ = 19460
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:

$\sqrt{\Delta}=\sqrt{19460}=\sqrt{4*4865}=\sqrt{4}*\sqrt{4865}=2\sqrt{4865}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-140)-2\sqrt{4865}}{2*35}=\frac{140-2\sqrt{4865}}{70}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-140)+2\sqrt{4865}}{2*35}=\frac{140+2\sqrt{4865}}{70}
$