Question

compute the value of the discriminant and give the number of real solutions of the quadratic equation.
−2x^2+6x-3=0

Answer 1

$-2x^2+6x-3=0 \\ \\ a=-2 \\ b=6 \\ c=-3 \\ \Delta=b^2-4ac=6^2-4 \times (-2) \times (-3)=36-24=12$

The value of the discriminant is 12.
Since the discriminant is greater than zero, the equation has two real solutions.