2m^2+19m+42=0

Solution for 2m^2+19m+42=0 equation:

2m^2+19m+42=0

a = 2; b = 19; c = +42;
Δ = b2-4ac
Δ = 192-4·2·42
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(19)-5}{2*2}=\frac{-24}{4}
=-6
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+5}{2*2}=\frac{-14}{4}
=-3+1/2
$