The given quadratic has real roots if its discriminant is non-negative, so the probability is
[tex]P(16Y^2-16(Y+2)\ge0)=P(16Y^2-16Y-32\ge0)[/tex]
[tex]=P(Y^2-Y-2\ge0)[/tex]
Complete the square:
[tex]Y^2-Y-2=Y^2-Y+\dfrac14-\dfrac94=\left(Y-\dfrac12\right)^2-\dfrac94[/tex]
[tex]\implies P(Y^2-Y-2\ge0)=P\left(\left(Y-\dfrac12\right)^2-\dfrac94\ge0\right)[/tex]
[tex]=P\left(\left(Y-\dfrac12\right)^2\ge\dfrac94\right)[/tex]
[tex]=P\left(\left|Y-\dfrac12\right|\ge\dfrac32\right)[/tex]
[tex]=P\left(Y-\dfrac12\ge\dfrac32\text{ and }Y-\dfrac12\ge-\dfrac32\right)[/tex]
[tex]=P\left(Y\ge2\text{ and }Y\ge-1\right)[/tex]
[tex]=P\left(Y\ge2\right)[/tex]
[tex]=\displaystyle\int_2^\infty f_Y(y)\,\mathrm dy[/tex]
where [tex]f_Y(y)[/tex] is the PDF of [tex]Y[/tex],
[tex]f_Y(y)=\begin{cases}\frac15&\text{for }0\le y\le5\\0&\text{otherwise}\end{cases}[/tex]
[tex]\implies P(Y\ge2)=\displaystyle\int_2^5\frac{\mathrm dy}5=\boxed{\frac35}[/tex]
