m^2=3m+1=0

Solution for m^2=3m+1=0 equation:

m^2=3m+1=0

We move all terms to the left:

m^2-(3m+1)=0

We get rid of parentheses

m^2-3m-1=0

a = 1; b = -3; c = -1;
Δ = b2-4ac
Δ = -32-4·1·(-1)
Δ = 13
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{-(-3)-\sqrt{13}}{2*1}=\frac{3-\sqrt{13}}{2}
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{13}}{2*1}=\frac{3+\sqrt{13}}{2}
$