Find all the values for a so that the piecewise function:

f(x): (x+1 when x≤a, x^2 when x>a is continuous.

Hello,

It bears repeating: you should have applied the well-known following formulas: "Hello", "Please" and "Thanks".
Anyway...

= = = = = = = = = = = = = = = = = =
For a given real value a, let function f be defined by:
   { ∀∈(-∞; a],    f(x) = x + 1
   { ∀∈(a; +∞),   f(x) = x²

This function is continuous iff:
   Lim(x→a⁻) f(x) = Lim(x→a⁺) f(x)
   Lim(x→a⁻) (x + 1) = Lim(x→a⁺) x²
   a + 1 = a²

So the function is continuous iff the chosen value "a" satisfies:
   a + 1 = a²
   a² – a – 1 = 0
   a² – 2×a×(½) – 1 = 0 →→→ Express as the beginning of a square
   a² – 2×a×(½) + (½)² – ¼ – 4/4 = 0 →→→ Complete the square
   (a – ½)² – (5/4) = 0 →→→ Because a²–2ab+b²=(a–b)²
   (a – ½)² – [(√5)/2]² = 0 →→→ Express remainder as a pertfect square
   [a – ½ – (√5)/2][a – ½ + (√5)/2] = 0 →→→ Because a²–b²=(a–b)(a+b)
   [a – (1 + √5)/2][a – (1 – √5)/2] = 0
And using null factor law,
   a = (1 ± √5)/2

So there are two values of a so that the piecewise function f will be continuous:
   a = (1 ± √5)/2

Regards,
Dragon.Jade :-)

***Sigh...***

"Do not give what is holy to the dogs; nor cast your pearls before swine, lest they trample them under their feet, and turn and tear you in pieces."
Matthew 7:6

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There are other methods beside factoring to solve a quadratic equation, quadratic formula for example or completing the square as shown below:
a^2 - a - 1 = 0
a^2 - a = 1
a^2 - a + (1/2)^2 = 1 + 1/4
(a - 1/2)^2 = 5/4
a - 1/2 = ±√5/2
a = 1/2(1 ± √5)

Please notice that above is the answer to your added question about the quadratics only.

Regards.

For the function to be continuous when x=a,

f(a) = a², i.e.
a+1 = a², i.e.
a² -a -1 =0

Now, solve for a.