3x^2+20=123

Solution for 3x^2+20=123 equation:

3x^2+20=123

We move all terms to the left:

3x^2+20-(123)=0

We add all the numbers together, and all the variables

3x^2-103=0

a = 3; b = 0; c = -103;
Δ = b2-4ac
Δ = 02-4·3·(-103)
Δ = 1236
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:

$\sqrt{\Delta}=\sqrt{1236}=\sqrt{4*309}=\sqrt{4}*\sqrt{309}=2\sqrt{309}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{309}}{2*3}=\frac{0-2\sqrt{309}}{6}
=-\frac{2\sqrt{309}}{6}
=-\frac{\sqrt{309}}{3}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{309}}{2*3}=\frac{0+2\sqrt{309}}{6}
=\frac{2\sqrt{309}}{6}
=\frac{\sqrt{309}}{3}
$