How do we know that the quadratic 3y2−y−12 has real root? (a) Notice the quadratic...

(a) This does not mean the quadratic has no real roots. A clearer example of this is
${\displaystyle y^2-2=(y+\sqrt{2})(y-\sqrt{2})}$.
(b) One way to know is to observe that plugging in ${\displaystyle y=0\text{and}y=3}$ yields
${\displaystyle 3\cdot 0^2-0-12=-12}$,
${\displaystyle 3\cdot 3^2-3-12=12}$.
so somewhere between 0 and 3 the quadratic must equal 0.
Another way to know is by computing the discriminant, which is
${\displaystyle \mathrm{\Delta }=b^2-4ac={(-1)}^2-4\cdot 3\cdot (-12)=145.}$
The quadratic has a real root because the discriminant is nonnegative.
(c) If r and s are roots of
${\displaystyle 3y^2-y-12,}$
then it follows that ${\displaystyle r+2\text{and}s+2}$ are roots of
${\displaystyle 3{(y-2)}^2-(y-2)-12}$
which by a little bit of algebra simplifies to
${\displaystyle 3y^2-13y+2}$.

Using Vieta theorem, ${\displaystyle r+s=\frac{1}{3},r\cdot s=-4}$.
Thus, ${\displaystyle (r+2)(s+2)=rs+2(r+s)+4=-4+\frac{2}{3}+4=\frac{2}{3}}$
and ${\displaystyle (r+2)+(s+2)=r+s+4=\frac{13}{3}}$
Applying Vieta in reverse, we can get a solution for c):
${\displaystyle 3y^2-13y+2=0}$
To show that there is a real root, rewrite the equation as ${\displaystyle 3{(x-\frac{1}{6})}^2=12+\frac{1}{12}}$ and use that any non-negative real can be written as a square of some real. Or you can use the hint provided and apply Intermediate Value Theorem.