3x^2+12x=14

Solution for 3x^2+12x=14 equation:

3x^2+12x=14

We move all terms to the left:

3x^2+12x-(14)=0

a = 3; b = 12; c = -14;
Δ = b2-4ac
Δ = 122-4·3·(-14)
Δ = 312
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:

$\sqrt{\Delta}=\sqrt{312}=\sqrt{4*78}=\sqrt{4}*\sqrt{78}=2\sqrt{78}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-2\sqrt{78}}{2*3}=\frac{-12-2\sqrt{78}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+2\sqrt{78}}{2*3}=\frac{-12+2\sqrt{78}}{6}
$