(7/8)b=4b+3

Solution for (7/8)b=4b+3 equation:

(7/8)b=4b+3

We move all terms to the left:

(7/8)b-(4b+3)=0


Domain of the equation: 8)b!=0

b!=0/1

b!=0

b∈R


We add all the numbers together, and all the variables

(+7/8)b-(4b+3)=0

We multiply parentheses

7b^2-(4b+3)=0

We get rid of parentheses

7b^2-4b-3=0

a = 7; b = -4; c = -3;
Δ = b2-4ac
Δ = -42-4·7·(-3)
Δ = 100
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{100}=10$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-10}{2*7}=\frac{-6}{14}
=-3/7
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+10}{2*7}=\frac{14}{14}
=1
$