The solution of quadratic equation
is
![x=2 \text { or } (-1)/(3)](https://img.qammunity.org/2020/formulas/mathematics/high-school/vjfkjeizzg98zc0b256yd3jp19skcnm43n.png)
Solution:
Given, equation is
![3 x^(2)-5 x-2=0](https://img.qammunity.org/2020/formulas/mathematics/high-school/ob13m31qf1o8htehfouancfy74utdigqan.png)
We have to solve the above given quadratic equation.
Now, take the given quadratic equation
![\rightarrow 3 x^(2)-5 x-2=0](https://img.qammunity.org/2020/formulas/mathematics/high-school/pyr7hztapwz6rtuzz8k5mqvx2xcw38yfkj.png)
Splitting “-5x” as -6x + x
![\rightarrow 3 x^(2)-6 x+x-2=0](https://img.qammunity.org/2020/formulas/mathematics/high-school/22q72lofv5u64iuyujvk1j8pmgph47j45b.png)
Take “3x” as common term from first two terms
![\rightarrow 3 x(x-2)+1(x-2)=0](https://img.qammunity.org/2020/formulas/mathematics/high-school/ghusz748p6jaq16aehcb51gj9ywplwpwup.png)
Take (x - 2) as common
![\rightarrow(x-2)(3 x+1)=0](https://img.qammunity.org/2020/formulas/mathematics/high-school/j9is5p164ljoi09p7yk986z51g2sk0ay2f.png)
Equating to zero we get,
![\begin{array}{l}{\rightarrow x-2=0 \text { or } 3 x+1=0} \\\\ {\rightarrow x=2 \text { or } 3 x=-1} \\\\ {\rightarrow x=2 \text { or } x=(-1)/(3)} \\\\ {\rightarrow x=2 \text { or } (-1)/(3)}\end{array}](https://img.qammunity.org/2020/formulas/mathematics/high-school/awus09sgousvrqia8uq73aciv0a9knx40b.png)
Hence, the roots the quadratic equation are 2 and
![(-1)/(3)](https://img.qammunity.org/2020/formulas/mathematics/college/fckhteizmd1nlxsrxqhazquv1xuql04xqg.png)