-3m^2+m+52=0

Solution for -3m^2+m+52=0 equation:

-3m^2+m+52=0

a = -3; b = 1; c = +52;
Δ = b2-4ac
Δ = 12-4·(-3)·52
Δ = 625
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{625}=25$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-25}{2*-3}=\frac{-26}{-6}
=4+1/3
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+25}{2*-3}=\frac{24}{-6}
=-4
$