3m^2+48m+23=0

Solution for 3m^2+48m+23=0 equation:

3m^2+48m+23=0

a = 3; b = 48; c = +23;
Δ = b2-4ac
Δ = 482-4·3·23
Δ = 2028
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{2028}=\sqrt{676*3}=\sqrt{676}*\sqrt{3}=26\sqrt{3}$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(48)-26\sqrt{3}}{2*3}=\frac{-48-26\sqrt{3}}{6}
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(48)+26\sqrt{3}}{2*3}=\frac{-48+26\sqrt{3}}{6}
$