3g^2+48g+105=0

Solution for 3g^2+48g+105=0 equation:

3g^2+48g+105=0

a = 3; b = 48; c = +105;
Δ = b2-4ac
Δ = 482-4·3·105
Δ = 1044
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$g_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$g_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{1044}=\sqrt{36*29}=\sqrt{36}*\sqrt{29}=6\sqrt{29}$
$g_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(48)-6\sqrt{29}}{2*3}=\frac{-48-6\sqrt{29}}{6}
$
$g_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(48)+6\sqrt{29}}{2*3}=\frac{-48+6\sqrt{29}}{6}
$