5m^2+44m+32=0

Solution for 5m^2+44m+32=0 equation:

5m^2+44m+32=0

a = 5; b = 44; c = +32;
Δ = b2-4ac
Δ = 442-4·5·32
Δ = 1296
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1296}=36$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(44)-36}{2*5}=\frac{-80}{10}
=-8
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(44)+36}{2*5}=\frac{-8}{10}
=-4/5
$