3x^2-24x+21=12

Solution for 3x^2-24x+21=12 equation:

3x^2-24x+21=12

We move all terms to the left:

3x^2-24x+21-(12)=0

We add all the numbers together, and all the variables

3x^2-24x+9=0

a = 3; b = -24; c = +9;
Δ = b2-4ac
Δ = -242-4·3·9
Δ = 468
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:

$\sqrt{\Delta}=\sqrt{468}=\sqrt{36*13}=\sqrt{36}*\sqrt{13}=6\sqrt{13}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-24)-6\sqrt{13}}{2*3}=\frac{24-6\sqrt{13}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-24)+6\sqrt{13}}{2*3}=\frac{24+6\sqrt{13}}{6}
$