b^2+3b=92+5b

Solution for b^2+3b=92+5b equation:

b^2+3b=92+5b

We move all terms to the left:

b^2+3b-(92+5b)=0

We add all the numbers together, and all the variables

b^2+3b-(5b+92)=0

We get rid of parentheses

b^2+3b-5b-92=0

We add all the numbers together, and all the variables

b^2-2b-92=0

a = 1; b = -2; c = -92;
Δ = b2-4ac
Δ = -22-4·1·(-92)
Δ = 372
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:

$\sqrt{\Delta}=\sqrt{372}=\sqrt{4*93}=\sqrt{4}*\sqrt{93}=2\sqrt{93}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-2\sqrt{93}}{2*1}=\frac{2-2\sqrt{93}}{2}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2\sqrt{93}}{2*1}=\frac{2+2\sqrt{93}}{2}
$