3=13b+6(b^2-2)

Solution for 3=13b+6(b^2-2) equation:

3=13b+6(b^2-2)

We move all terms to the left:

3-(13b+6(b^2-2))=0


We calculate terms in parentheses: -(13b+6(b^2-2)), so:

13b+6(b^2-2)

We multiply parentheses

6b^2+13b-12

Back to the equation:

-(6b^2+13b-12)


We get rid of parentheses

-6b^2-13b+12+3=0

We add all the numbers together, and all the variables

-6b^2-13b+15=0

a = -6; b = -13; c = +15;
Δ = b2-4ac
Δ = -132-4·(-6)·15
Δ = 529
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{529}=23$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-13)-23}{2*-6}=\frac{-10}{-12}
=5/6
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-13)+23}{2*-6}=\frac{36}{-12}
=-3
$