3x^2+12=63

Solution for 3x^2+12=63 equation:

3x^2+12=63

We move all terms to the left:

3x^2+12-(63)=0

We add all the numbers together, and all the variables

3x^2-51=0

a = 3; b = 0; c = -51;
Δ = b2-4ac
Δ = 02-4·3·(-51)
Δ = 612
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:

$\sqrt{\Delta}=\sqrt{612}=\sqrt{36*17}=\sqrt{36}*\sqrt{17}=6\sqrt{17}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-6\sqrt{17}}{2*3}=\frac{0-6\sqrt{17}}{6}
=-\frac{6\sqrt{17}}{6}
=-\sqrt{17}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+6\sqrt{17}}{2*3}=\frac{0+6\sqrt{17}}{6}
=\frac{6\sqrt{17}}{6}
=\sqrt{17}
$