Question 259039
a)
2x^2 - kx - 3=0
just plug in 3 for x and solve for k
2*3^2 -3k - 3=0
2*9-3k-3=0
15=3k
5=k
b)
b^2-4ac must =0
(-1)^2-4(k*k)=0
1=4k^2
1/4=k^2
k=+1/2 and -1/2
(-1)^2-4*(-1/2)(-1/2
1-1=0
also works with +1/2
c)
2x^2 - x - 3 = -5
2x^2-x+2=0

 two complex numbers because the discriminant(b^2-4ac=(-1)^2-(4*2*(-3)) is less than zero

*[invoke quadratic "x", 2, -1, 2 ]