3b^2-19b-30=-b^2

Solution for 3b^2-19b-30=-b^2 equation:

3b^2-19b-30=-b^2

We move all terms to the left:

3b^2-19b-30-(-b^2)=0

We get rid of parentheses

3b^2+b^2-19b-30=0

We add all the numbers together, and all the variables

4b^2-19b-30=0

a = 4; b = -19; c = -30;
Δ = b2-4ac
Δ = -192-4·4·(-30)
Δ = 841
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{841}=29$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-19)-29}{2*4}=\frac{-10}{8}
=-1+1/4
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-19)+29}{2*4}=\frac{48}{8}
=6
$