9x^2=21x+6

Solution for 9x^2=21x+6 equation:

9x^2=21x+6

We move all terms to the left:

9x^2-(21x+6)=0

We get rid of parentheses

9x^2-21x-6=0

a = 9; b = -21; c = -6;
Δ = b2-4ac
Δ = -212-4·9·(-6)
Δ = 657
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:

$\sqrt{\Delta}=\sqrt{657}=\sqrt{9*73}=\sqrt{9}*\sqrt{73}=3\sqrt{73}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-21)-3\sqrt{73}}{2*9}=\frac{21-3\sqrt{73}}{18}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-21)+3\sqrt{73}}{2*9}=\frac{21+3\sqrt{73}}{18}
$