The roots of the quadratic equation 2$$x^2$$​-x-3=0 are:

Correct option is B

Given: 

​$$2x^2-x-3=0$$ 

Formula used:

Discrimanant (D) = $$b^2 - 4ac$$​​

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

Solution:

Here : $$a=1 , b = -1, c= -3$$ 

D = $$b^2 - 4ac$$  = $$(−1) 2 −4(2)(−3)=1+24=25$$

Substitute the value in above formula:

​$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ = $$\frac{-(-1) \pm \sqrt{25}}{2(2)} = \frac{1 \pm 5}{4}$$ 

Solve the root:

​$$x_1 = \frac{1 + 5}{4} = \frac{6}{4} = \frac{3}{2}$$ 

$$x_2 = \frac{1 - 5}{4} = \frac{-4}{4} = -1$$ 

So the root of quadratic equation is  $$\frac{3}{2}, -1$$

Thus the correct answer is (B)