Question

Compute the value of the discriminant and give the number of real solutions of the quadratic equation.-2x² + 6x - 3=0Discriminant:12Number of real solutions:

Answer 1

Given:

$-2x^2+6x-3=0$$a=-2\text{ ; b=6 ; c=-3}$$D=b^2-4ac$$D=6^2-4(-2)(-3)$$D=36-24$$D=12$

Discriminant = 12

12>0

Therefore, the quadratic equation has two real roots.