How many real numbers solutions are there to the equation?

0 = - 3x^2+x-4

ax^2+bx+c=0

The quadratic formula is

x = (-b + / - √ (b^2-4ac)) / 2a

The part √ (b^2-4ac) tells us how many solution we will get:

using - 3x^2+x-4

√ (1-4 (-4) (-3)) = √ (1-36) = √-35 this is an impossible result since the inside of a square root has to be always greater or equal to zero. This means that the equation has no solutions.