m1+3m2=500

Solution for m1+3m2=500 equation:

m1+3m^2=500

We move all terms to the left:

m1+3m^2-(500)=0

We add all the numbers together, and all the variables

3m^2+m-500=0

a = 3; b = 1; c = -500;
Δ = b2-4ac
Δ = 12-4·3·(-500)
Δ = 6001
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-\sqrt{6001}}{2*3}=\frac{-1-\sqrt{6001}}{6}
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{6001}}{2*3}=\frac{-1+\sqrt{6001}}{6}
$