Answering! Was just making sure I could actually post because it was giving me trouble the other day. Editing now.

Turn the word problem into an equation.  Call the unknown number x. The square of that number is x². The square of 5 more than that number is (x+5)². The sum of those two quantities, x² + (x+5)², equals 157.

OK, let's solve it.

x² + (x+5)² = 157

x² + (x² + 10x + 25) = 157

2x² + 10x + 25 = 157

2x² + 10x - 132 = 0

This is the standard quadratic form, ax² + bx + c = 0.  There will be two solutions.  You can solve the equation by factoring, or by the quadratic formula.  If I don't see an easy factorization pattern right away, I just go straight to the quadratic formula myself.

x = [-b ± √(b² - 4ac) ] / 2a

x = [-10 ± √(10² - 4 · 2 · -132) ] / (2 · 2)

x = [-10 ± √(100+1056) ] / 4

x = [-10 ± √1156] / 4

x = [-10 ± 34] / 4

x = -44 / 4 , +24 / 4

x = -11, 6

But the problem tells us one more thing: x is positive.  Therefore, x = 6.

x = First number

x + 5 = Second number

x² + (x + 5)² = 157                       Equation derived from problem

x² + x² + 10x + 25 = 157              Square the binomial (FOIL)

2x² + 10x + 25 = 157                   Simplify

2x² + 10x + 25 - 157 = 157 - 157  Subtract 157 from each side

2x² + 10x - 132 = 0                      Simplify

(2x² + 10x - 132)/2 = 0/2             Divide each side by 2

x² + 5x - 66 = 0                           Simplify

(x + 11)(x - 6) = 0                       Factor

x + 11 = 0 or x - 6 = 0                Set each factor equal to 0

x = -11 or x = 6                        Solve each factor for x

The answer is 6, because you are looking for the positive number.

Let's call the number x. 

x² + (x + 5)² = 157

x² + (x² + 10x + 25) = 157

2x² + 10x + 25 - 157 = 157 - 157

2x² + 10x - 132 = 0 now divide both sides by 2

x² + 5x - 66 = 0

You can use the quadratic formula here, but it looks like we can factor this. We want to find two factors that multiply to equal -66 and add up to 5. They are 11 and -6, so we can write 

(x + 11)(x - 6) = 0

The number has to be positive, so ignore x + 11 = 0 as that'll give us a negative

x - 6 = 0

x = 6

Now we can check that 6 works in the original equation...

x² + (x + 5)² = 157

6² + (6 + 5)² = 157

36 + 121 = 157

the sum of the square of a positive number and the square of 5 more than the number is 157. what is the number?

Let n = the positive number

then

from: "the sum of the square of a positive number and the square of 5 more than the number is 157" we get

n^2 + (n+5) ^2 = 157

n^2 + (n+5)(n+5) = 157

n^2 + n^2+10n+25 = 157

2n^2+10n+25 = 157

2n^2+10n-132 = 0

n^2+5n-66 = 0

(n+11)(n-6) = 0

n ={-11, 6}

since we're looking for a POSITIVE number, throw out the -11 leaving

n = 6

Let us call the unknown number "n". The sum of (the square of a positive number) PLUS (the square of 5 more than the number is equal to 157. Mathematically, this can be represented as:

n² + (n + 5)² = 157

Expanding this, we get:

n² + (n + 5)(n + 5) = 157

which equals:

n² + n² + 10n + 25 = 157

or:

2n² + 10n - 132 = 0

We can divide each side of the equation by 2 and get the following:

n² + 5n - 66 = 0

Factoring this, we get (n + 11)(n - 6) = 0, and n can equal -11 or 6. However, because the original question states that the unknown "n" is positive, we know that it must therefore be 6 and not -11.