7m^2+53m+28=0

Solution for 7m^2+53m+28=0 equation:

7m^2+53m+28=0

a = 7; b = 53; c = +28;
Δ = b2-4ac
Δ = 532-4·7·28
Δ = 2025
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2025}=45$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(53)-45}{2*7}=\frac{-98}{14}
=-7
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(53)+45}{2*7}=\frac{-8}{14}
=-4/7
$