49m^2+16m=0

Solution for 49m^2+16m=0 equation:

49m^2+16m=0

a = 49; b = 16; c = 0;
Δ = b2-4ac
Δ = 162-4·49·0
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-16}{2*49}=\frac{-32}{98}
=-16/49
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+16}{2*49}=\frac{0}{98}
=0
$