3x^2+12x+3=22

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

3x^2+12x+3=22

We move all terms to the left:

3x^2+12x+3-(22)=0

We add all the numbers together, and all the variables

3x^2+12x-19=0

a = 3; b = 12; c = -19;
Δ = b2-4ac
Δ = 122-4·3·(-19)
Δ = 372
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:

$\sqrt{\Delta}=\sqrt{372}=\sqrt{4*93}=\sqrt{4}*\sqrt{93}=2\sqrt{93}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-2\sqrt{93}}{2*3}=\frac{-12-2\sqrt{93}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+2\sqrt{93}}{2*3}=\frac{-12+2\sqrt{93}}{6}
$