Part 1
We are given [tex]x^2-21x=-4x[/tex]. This can be rewritten as [tex]x^2-18x=0[/tex].
Therefore, a=1, b=-18, c=0.
Using the quadratic formula
[tex]x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}=\frac{-\left(-18\right)\pm \sqrt{\left(-18\right)^2-4\left(1\right)\left(0\right)}}{2\left(1\right)}[/tex]
[tex]x=\frac{18\pm 18}{2}[/tex]
The values of x are
[tex]x_1=\frac{18-18}{2}=0[/tex]
[tex]x_2=\frac{18+18}{2}=18[/tex]
Part 2
Since the values of y change drastically for every equal interval of x, the function cannot be linear. Therefore, the kind of function that best suits the given pairs is a quadratic function.
Part 3.
The first equation is [tex]y=x^2+2[/tex].
The second equation is [tex]y=3x+20[/tex].
We have
[tex]x^2+2=3x+20[/tex]
[tex]x^2-3x-18=0[/tex]
Factoring, we have
[tex]\left(x-6\right)\left(x+3\right)=0[/tex]
Equating both factors to zero.
[tex]x_1-6=0\rightarrow x_1=6[/tex]
[tex]x_2+3=0\rightarrow x_2=-3[/tex]
When the value of x is 6, the value of y is
[tex]y=3\left(6\right)+20=38[/tex]
When the value of x is -3, the value of y is
[tex]y=3\left(-3\right)+20=11[/tex]
Therefore, the solutions are (6,38) or (-3,11)
