3.x^2+12=48

Solution for 3.x^2+12=48 equation:

3.x^2+12=48

We move all terms to the left:

3.x^2+12-(48)=0

We add all the numbers together, and all the variables

3.x^2-36=0

a = 3.; b = 0; c = -36;
Δ = b2-4ac
Δ = 02-4·3.·(-36)
Δ = 432
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:

$\sqrt{\Delta}=\sqrt{432}=\sqrt{144*3}=\sqrt{144}*\sqrt{3}=12\sqrt{3}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-12\sqrt{3}}{2*3.}=\frac{0-12\sqrt{3}}{6}
=-\frac{12\sqrt{3}}{6}
=-2\sqrt{3}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+12\sqrt{3}}{2*3.}=\frac{0+12\sqrt{3}}{6}
=\frac{12\sqrt{3}}{6}
=2\sqrt{3}
$