. Find the nature of the roots of the following quadratic equations. If the real roots exist, find them; 2x^2 - 6x + 3 = 0

Dear Student

2x^2- 6x+ 3 = 0

Comparing the equation withax^2+bx+c= 0, we get

a= 2,b= -6,c= 3

As we know, Discriminant =b^2- 4ac

= (-6)^2- 4 (2) (3)
= 36 - 24 = 12
b^2-4ac>0,
Therefore, there are distinct real roots exist for this equation,
2x^2- 6x+ 3 = 0.

x = (-(-6) ±√(-6^2-4(2)(3)) )/ 2(2)
= (6±2√3 )/4
= (3±√3)/2
Therefore the roots for the given equation are(3+√3)/2 and (3-√3)/2

Thanks