3b2=-b+24

Solution for 3b2=-b+24 equation:

3b^2=-b+24

We move all terms to the left:

3b^2-(-b+24)=0

We add all the numbers together, and all the variables

3b^2-(-1b+24)=0

We get rid of parentheses

3b^2+1b-24=0

We add all the numbers together, and all the variables

3b^2+b-24=0

a = 3; b = 1; c = -24;
Δ = b2-4ac
Δ = 12-4·3·(-24)
Δ = 289
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{289}=17$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-17}{2*3}=\frac{-18}{6}
=-3
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+17}{2*3}=\frac{16}{6}
=2+2/3
$