For this case we have that by definition, the roots, or also called zeros, of the quadratic function are those values of x for which the expression is 0.
Then, we must find the roots of:

Where:

We have to:
Substituting we have:

By definition we have to:

So:

Thus, we have two roots:

Answer:

Step-by-step explanation:
0 is neither a rational number nor an irrational number
Sqrt (y^4 * y)
Sqrt y*4 = y^2
y^2 sqrt y
Y-y1= m(x-x1) point-slope form
y-0= -3(x-5)
solve this point-slope form to slope-intercept form. Use the information to graph the line.