17m^2+19m+2=0

Solution for 17m^2+19m+2=0 equation:

17m^2+19m+2=0

a = 17; b = 19; c = +2;
Δ = b2-4ac
Δ = 192-4·17·2
Δ = 225
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{225}=15$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(19)-15}{2*17}=\frac{-34}{34}
=-1
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+15}{2*17}=\frac{-4}{34}
=-2/17
$