the sum of the squares of two consecutive integers is 365. find the integers using the quadratic formula.

if possible, show your steps! :)

a^2+ (a+1)^2=365

a^2 + a^2+2a+1=365
2a^2+2a-364=0
a^2+a-182=0
now apply the quadratic equation to find the a, the lower number. In my head, I get somehing in the lower teens

To find the consecutive integers, we can set up an equation based on the given information.

Let's assume the first integer is "x". Since the consecutive integer will be the next number, we can label it as "x + 1".

According to the given information, the sum of the squares of these integers is 365. Therefore, our equation becomes:

x^2 + (x + 1)^2 = 365

Expanding the equation:

x^2 + (x + 1)(x + 1) = 365
x^2 + (x^2 + 2x + 1) = 365
2x^2 + 2x + 1 = 365

Rearranging the equation:

2x^2 + 2x + 1 - 365 = 0
2x^2 + 2x - 364 = 0

Now that we have a quadratic equation in the form of "ax^2 + bx + c = 0," we can apply the quadratic formula:

x = [-b ± √(b^2 - 4ac)] / (2a)

For our equation, a = 2, b = 2, and c = -364. Plugging these values into the quadratic formula:

x = [-2 ± √(2^2 - 4(2)(-364))] / (2(2))
x = [-2 ± √(4 + 2912)] / 4
x = [-2 ± √(2916)] / 4

Simplifying further:

x = [-2 ± 54] / 4

Now, by solving for both the positive and negative values of x using the quadratic formula, we will get the two possible values for the first integer:

1) When x = (-2 + 54) / 4
x = 52 / 4
x = 13

2) When x = (-2 - 54) / 4
x = -56 / 4
x = -14

Therefore, the two possible pairs of consecutive integers whose squares sum up to 365 are (13, 14) and (-14, -13).

Note: It's important to note that we obtained both positive and negative values for x, as quadratic equations can have two solutions, unless restricted by the context of the problem. In this case, considering both possibilities allows us to find both pairs of consecutive integers that satisfy the given condition.