-40=3x^2+23x

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

-40=3x^2+23x

We move all terms to the left:

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

We get rid of parentheses

-3x^2-23x-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:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

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