3x^2+22x+4=0

Solution for 3x^2+22x+4=0 equation:

3x^2+22x+4=0

a = 3; b = 22; c = +4;
Δ = b2-4ac
Δ = 222-4·3·4
Δ = 436
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:

$\sqrt{\Delta}=\sqrt{436}=\sqrt{4*109}=\sqrt{4}*\sqrt{109}=2\sqrt{109}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(22)-2\sqrt{109}}{2*3}=\frac{-22-2\sqrt{109}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(22)+2\sqrt{109}}{2*3}=\frac{-22+2\sqrt{109}}{6}
$