3x^2+12x+19=17

Solution for 3x^2+12x+19=17 equation:

3x^2+12x+19=17

We move all terms to the left:

3x^2+12x+19-(17)=0

We add all the numbers together, and all the variables

3x^2+12x+2=0

a = 3; b = 12; c = +2;
Δ = b2-4ac
Δ = 122-4·3·2
Δ = 120
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:

$\sqrt{\Delta}=\sqrt{120}=\sqrt{4*30}=\sqrt{4}*\sqrt{30}=2\sqrt{30}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-2\sqrt{30}}{2*3}=\frac{-12-2\sqrt{30}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+2\sqrt{30}}{2*3}=\frac{-12+2\sqrt{30}}{6}
$