0=690-212x+12x^2

Solution for 0=690-212x+12x^2 equation:

0=690-212x+12x^2

We move all terms to the left:

0-(690-212x+12x^2)=0

We add all the numbers together, and all the variables

-(690-212x+12x^2)=0

We get rid of parentheses

-12x^2+212x-690=0

a = -12; b = 212; c = -690;
Δ = b2-4ac
Δ = 2122-4·(-12)·(-690)
Δ = 11824
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:

$\sqrt{\Delta}=\sqrt{11824}=\sqrt{16*739}=\sqrt{16}*\sqrt{739}=4\sqrt{739}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(212)-4\sqrt{739}}{2*-12}=\frac{-212-4\sqrt{739}}{-24}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(212)+4\sqrt{739}}{2*-12}=\frac{-212+4\sqrt{739}}{-24}
$