A quadratic equation whose roots are $$\frac{-2}3$$​ and $$\frac{-3}2$$​ is:

Correct option is C

Given:

Roots of quadratic equation  = $$\frac{-2}3$$​ and  $$\frac{-3}2$$​​

Concept Used: 

For a quadratic equation of the form:  $$ax^2 + bx + c = 0$$ 

Sum of the roots = α + β  =  $$-\frac{b}{a}$$​​

Product of the roots = α β  = $$\frac{c}{a}$$​​

Solution: 

the roots of the quadratic equation are $$\alpha = -\frac{2}{3}$$​ and $$\beta = -\frac{3}{2}$$​  

sum of the roots: $$\alpha + \beta = \frac{-2}{3} + \frac{-3}{2} = \frac{(-4-9)}{6}= -\frac{13}{6}$$ 

product of the roots:  $$\alpha\cdot\beta = \frac{-2}{3} \times \frac{-3}{2} = \frac{6}{6} = 1$$

​​Now, using the standard form $$x^2 - (\alpha + \beta) x + \alpha \beta = 0$$ :

$$x^2 - \left( \frac{-13}{6} \right) x + 1 = 0$$ 

$$x^2 + \frac{13}{6}x + 1 = 0$$ 

$$6x^2 + 13x + 6 = 0$$ 

​Thus, the quadratic equation whose roots are $$-\frac{2}{3}$$ and $$-\frac{3}{2}$$​ is $$6x^2 + 13x + 6 = 0$$

​​