3b^2+40=-23b

Solution for 3b^2+40=-23b equation:

3b^2+40=-23b

We move all terms to the left:

3b^2+40-(-23b)=0

We get rid of parentheses

3b^2+23b+40=0

a = 3; b = 23; c = +40;
Δ = b2-4ac
Δ = 232-4·3·40
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(23)-7}{2*3}=\frac{-30}{6}
=-5
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(23)+7}{2*3}=\frac{-16}{6}
=-2+2/3
$