Question

How many real number solutions are there to the equation 0= -3x^2+x-4?
0
1
2
3

Answer 1

-3x^2+x-4=0
3x^2-x+4=0
x=\[(1+-\sqrt{1^2-4*4*3})\]

Answer 2

divide by 2*3

Answer 3

for a quadratic
ax2+bx+c=0
use the discriminant
Δ=b2−4ac
in your question a=-3, b = 1 and c= -4 then 1^2 - 4 x (-3) x (-4) = -47 since
Δ<0
no real solutions Rules
Δ>0
2 unequal real solutions if
Δ
is a square number the roots are rational.
Δ=0
repeated or equal roots and the senario above

Answer 4

as quantity in root is negative so 0 is the number of real solution

Answer 5

Thanks