3x^2+22x+12=0

Solution for 3x^2+22x+12=0 equation:

3x^2+22x+12=0

a = 3; b = 22; c = +12;
Δ = b2-4ac
Δ = 222-4·3·12
Δ = 340
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:

$\sqrt{\Delta}=\sqrt{340}=\sqrt{4*85}=\sqrt{4}*\sqrt{85}=2\sqrt{85}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(22)-2\sqrt{85}}{2*3}=\frac{-22-2\sqrt{85}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(22)+2\sqrt{85}}{2*3}=\frac{-22+2\sqrt{85}}{6}
$