Question

Solve the quadratic equation by quadratic formula .$\frac{2}{3}{q}^2+\frac{3}{10}q=\frac{1}{10}$

Short answer

The required result is$q=\frac{-9+\sqrt{321}}{40},\frac{-9-\sqrt{321}}{40}$

Step-by-step solution

 Given Information 

The given equation is$\frac{2}{3}{q}^2+\frac{3}{10}q=\frac{1}{10}$

Simplify the expression `to write in standard form  

LCM,

$$\begin{gathered} \frac{20q^2+9q}{30}=\frac{1}{10} \\ \frac{20q^2+9q}{3}=1 \\ 20q^2+9q=3 \end{gathered}$$

$20q^2+9q-3=0$

After converting in standard form .Identify the value of  a,b,c

$a=20,b=9,c=-3$

 Substitute the values in quadratic formula.  

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$$\begin{gathered} q=\frac{-\left(9\right)\pm \sqrt{{\left(9\right)}^2-4\left(20\right)\left(-3\right)}}{2\left(20\right)} \\ q=\frac{-9\pm \sqrt{81+240}}{40} \\ q=\frac{-9\pm \sqrt{321}}{40} \end{gathered}$$

$q=\frac{-9+\sqrt{321}}{40},\frac{-9-\sqrt{321}}{40}$