2m^2+18m+28=0

Solution for 2m^2+18m+28=0 equation:

2m^2+18m+28=0

a = 2; b = 18; c = +28;
Δ = b2-4ac
Δ = 182-4·2·28
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{-(18)-10}{2*2}=\frac{-28}{4}
=-7
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+10}{2*2}=\frac{-8}{4}
=-2
$