Question

Solve the equation using quadratic formula.

$6z^2-9z+1=0$

Short answer

The roots are$z=\frac{9+\sqrt{57}}{12},z=\frac{9-\sqrt{57}}{12}$.

Step-by-step solution

 Given Information.  

The given equation is $6z^2-9z+1=0$.

Write the quadratic equation in standard form and identify  a,b,c.

The quadratic equation is in standard form.

So,

$a=6,b=-9,c=1$

Substitute the values in quadratic formula.  

$$\begin{gathered} z=\frac{-b\pm \sqrt{b^2-4ac}}{2a} \\ z=\frac{-(-9)\pm \sqrt{(-9{)}^2-4(6\left)\right(1)}}{2\left(6\right)} \\ z=\frac{9\pm \sqrt{81-24}}{12} \\ z=\frac{9\pm \sqrt{57}}{12} \end{gathered}$$

Rewrite to show two solutions.  

$$\begin{gathered} z=\frac{9+\sqrt{57}}{12} \\ z=\frac{9-\sqrt{57}}{12} \end{gathered}$$

Thus, the roots are$z=\frac{9+\sqrt{57}}{12},z=\frac{9-\sqrt{57}}{12}$.