3p^2+29p+40=0

Solution for 3p^2+29p+40=0 equation:

3p^2+29p+40=0

a = 3; b = 29; c = +40;
Δ = b2-4ac
Δ = 292-4·3·40
Δ = 361
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{361}=19$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(29)-19}{2*3}=\frac{-48}{6}
=-8
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(29)+19}{2*3}=\frac{-10}{6}
=-1+2/3
$