3x2+12=312

Solution for 3x2+12=312 equation:

3x^2+12=312

We move all terms to the left:

3x^2+12-(312)=0

We add all the numbers together, and all the variables

3x^2-300=0

a = 3; b = 0; c = -300;
Δ = b2-4ac
Δ = 02-4·3·(-300)
Δ = 3600
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{3600}=60$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-60}{2*3}=\frac{-60}{6}
=-10
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+60}{2*3}=\frac{60}{6}
=10
$