Answer: (a) \(3\left(x+\frac52\right)^2-\frac{155}{4}\); (b) minimum value \(-\frac{155}{4}\) when \(x=-\frac52\); (c) \(y=1.31\) or \(y=-226\).

Answer: (a) \(3\left(x+\frac52\right)^2-\frac{155}{4}\); (b) minimum value \(-\frac{155}{4}\) when \(x=-\frac52\); (c) \(y=1.31\) or \(y=-226\).

(a) Complete the square:

\(3x^2+15x-20=3(x^2+5x)-20.\)

Now

\(x^2+5x=\left(x+\frac52\right)^2-\frac{25}{4}.\)

Therefore

\(3x^2+15x-20=3\left(x+\frac52\right)^2-\frac{75}{4}-20.\)

So

\(3x^2+15x-20=3\left(x+\frac52\right)^2-\frac{155}{4}.\)

(b) Since \(3\left(x+\frac52\right)^2\geq0\), the minimum value is

\(-\frac{155}{4},\)

and it occurs when

\(x=-\frac52.\)

(c) Let

\(u=y^{1/3}.\)

Then \(y^{2/3}=u^2\), so the equation becomes

\(3u^2+15u-20=0.\)

Using the completed square form,

\(3\left(u+\frac52\right)^2-\frac{155}{4}=0.\)

Thus

\(3\left(u+\frac52\right)^2=\frac{155}{4},\)

so

\(u=-\frac52\pm\sqrt{\frac{155}{12}}.\)

Since \(u=y^{1/3}\),

Therefore

\(y=1.31\quad\text{or}\quad y=-226\)

to 3 significant figures.