3m^2+22m+32=0

Solution for 3m^2+22m+32=0 equation:

3m^2+22m+32=0

a = 3; b = 22; c = +32;
Δ = b2-4ac
Δ = 222-4·3·32
Δ = 100
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{100}=10$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(22)-10}{2*3}=\frac{-32}{6}
=-5+1/3
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(22)+10}{2*3}=\frac{-12}{6}
=-2
$