3a2+12a=14

Solution for 3a2+12a=14 equation:

3a^2+12a=14

We move all terms to the left:

3a^2+12a-(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:
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

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