Quadratic equation $9x^2-37=6x$ using the quadratic formula

Question

Quadratic equation using the quadratic formula $9x^2-37=6x$

So $9x^2-6x-37=0$

$A= 9$ $b=-6$ $c=37$

$\dfrac{-(-6) \pm \sqrt{ (-6)^2- 4(9)(37)}}{2(9)}$, $\dfrac{6 \pm \sqrt{36-1332}}{18}$, $\dfrac{6 \pm \sqrt {1296}}{18}$

Kindly check if this is right? And what's next?

Answer 1

The solution of the equation $9x^2-6x-37=0$ is $$ x=\frac{-(-6)\pm\sqrt{(-6)^2-4\cdot 9\cdot (-37)}}{2\cdot 9}=\frac{6\pm\sqrt{36+1332}}{18}=\frac{6\pm\sqrt{1368}}{18} $$

Answer 2

$9x^2-6x-37 = 0$

$$\begin{align} x & = \dfrac{-(-6) \pm \sqrt{(-6)^2 - 4(9)(-37)}}{2(9)} \\ & = \dfrac{6 \pm \sqrt{36 + 1332}}{18} \\ & = \dfrac{6 \pm \sqrt{1368}}{18} \\ & = \dfrac{6 \pm 6\sqrt{38}}{18} \\ & = \dfrac{1 \pm \sqrt{38}}{3} \\ \end{align}$$

So $x \approx -1.721$ or $x \approx 2.388$

Answer 3

If you know about completing the square, you could also notice that $9x^2-6x+1=(3x-1)^2$.

Therefore your equation is equivalent to:
$9x^2-6x-37=0$
$(3x-1)^2-38=0$
$(3x-1)^2=38$

If you denote (substitute) $y=3x-1$, you get the equation $y^2=38$, which has two solutions $\pm\sqrt{38}$.

So by going back to the original equation, you get
$3x-1=\pm\sqrt{38}$
$3x=1\pm\sqrt{38}$
$x=\frac{1\pm\sqrt{38}}3$

You should get the same result if you try to simplify your result from quadratic formula a bit.