b(b^-4)=b^-3

Solution for b(b^-4)=b^-3 equation:

b(b^-4)=b^-3

We move all terms to the left:

b(b^-4)-(b^-3)=0

We multiply parentheses

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

We get rid of parentheses

b^2-4b-b^+3=0

We add all the numbers together, and all the variables

b^2-5b+3=0

a = 1; b = -5; c = +3;
Δ = b2-4ac
Δ = -52-4·1·3
Δ = 13
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-\sqrt{13}}{2*1}=\frac{5-\sqrt{13}}{2}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+\sqrt{13}}{2*1}=\frac{5+\sqrt{13}}{2}
$