3x^2+42x+3=0

Solution for 3x^2+42x+3=0 equation:

3x^2+42x+3=0

a = 3; b = 42; c = +3;
Δ = b2-4ac
Δ = 422-4·3·3
Δ = 1728
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{1728}=\sqrt{576*3}=\sqrt{576}*\sqrt{3}=24\sqrt{3}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(42)-24\sqrt{3}}{2*3}=\frac{-42-24\sqrt{3}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(42)+24\sqrt{3}}{2*3}=\frac{-42+24\sqrt{3}}{6}
$