If the discriminant is negative how many solutions are there

  no real solutions

Step-by-step explanation:

If it is a quadratic equation, there are always two solutions, regardless of the value of the discriminant.

When the discriminant is negative there are no real solutions, but there are two complex solutions.

no not all of them.

step-by-step explanation:

some of them have a square base.

step-by-step explanation:

i think something is missing in your question because y-1=(x-12) would not pass through the point (-4,-3). because if you replace x by -4 your are supposed to find y=-3 which is not the case her. if you replace x by -4 you find =-15.

i assume the right equation is y-1=1/4*(x-12) because this one is passing through both your points (if you are interested to know how i found that out i can tell you).

in this equation if i replace x by -4 i find y=-3 and if i replace x by 12, i find y=1 so the line representing this equation is passing through those 2 points.

then the standard form of the equation is finding an expression of y in function of x

y-1=1/4*(x-12)

y=x/4-3+1

y=x/4-2

hope that .

do you confirm there was a mistake in the question?