3x^2+9=21x

Solution for 3x^2+9=21x equation:

3x^2+9=21x

We move all terms to the left:

3x^2+9-(21x)=0

a = 3; b = -21; c = +9;
Δ = b2-4ac
Δ = -212-4·3·9
Δ = 333
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:

$\sqrt{\Delta}=\sqrt{333}=\sqrt{9*37}=\sqrt{9}*\sqrt{37}=3\sqrt{37}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-21)-3\sqrt{37}}{2*3}=\frac{21-3\sqrt{37}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-21)+3\sqrt{37}}{2*3}=\frac{21+3\sqrt{37}}{6}
$