3b^2=-b+140

Solution for 3b^2=-b+140 equation:

3b^2=-b+140

We move all terms to the left:

3b^2-(-b+140)=0

We add all the numbers together, and all the variables

3b^2-(-1b+140)=0

We get rid of parentheses

3b^2+1b-140=0

We add all the numbers together, and all the variables

3b^2+b-140=0

a = 3; b = 1; c = -140;
Δ = b2-4ac
Δ = 12-4·3·(-140)
Δ = 1681
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1681}=41$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-41}{2*3}=\frac{-42}{6}
=-7
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+41}{2*3}=\frac{40}{6}
=6+2/3
$