2 sides of an equilateral triangle measure (y+10) units and (y^2-2) units. If the perimeter of the triangle is 21 units, what is the value of y?

I was trying (y+(y+10) + (y^2-2)=21
got to y^2+2y+-13 and wasn't sure how to proceed.
Or should I take a side (y^2-2)*3=21?
I ended up with 3 but the other side ends up with -3. Could this be a typo?

You know the perimeter is 21 and, it's equilateral. Which means all sides are equal so therefor 21 divided by 3 = 7.

It appears that the 3rd side is y, so yes, that means

y + (y+10) + (y^2-2) = 21
y^2+2y+8 = 21
y^2+2y-13 = 0

now use the quadratic formula. If you want an integer or rational answer, there must be a typo somewhere.

To find the value of y, let's consider the sum of the three sides of the equilateral triangle:

(y + 10) + (y^2 - 2) + (y + 10) = 21

Simplifying this equation, we have:

2y + y^2 + 18 = 21

Rearranging the terms, we get:

y^2 + 2y - 3 = 0

To solve this quadratic equation, we can factor it as:

(y + 3)(y - 1) = 0

Setting each factor equal to zero gives us two possible solutions:

y + 3 = 0 -> y = -3
or
y - 1 = 0 -> y = 1

Since the side lengths cannot be negative, we can conclude that y = 1.

To solve this problem, we need to set up an equation based on the given information and then solve for the value of y.

Let's start by using the equation for the perimeter of an equilateral triangle, which states that the sum of all three sides is equal to the perimeter. In this case, the perimeter is given as 21 units.

So, the equation can be written as:
(y + 10) + (y^2 - 2) + (y + 10) = 21

Simplifying the equation:
y + 10 + y^2 - 2 + y + 10 = 21
y^2 + 2y + 28 = 21

Now, we need to solve for y. Rearrange the equation to set it equal to zero:
y^2 + 2y + 28 - 21 = 0
y^2 + 2y + 7 = 0

To further factorize this quadratic equation, we need to find two values whose sum is 2 and product is 7. Since there are no such two values, we can conclude that this equation cannot be factored. In this case, we can use the quadratic formula to find the value of y.

The quadratic formula states that for an equation in the form of ax^2 + bx + c = 0, the solutions for x can be found using the following formula:

x = (-b ± √(b^2 - 4ac)) / (2a)

Comparing this formula to our equation, we have a = 1, b = 2, and c = 7.

Using the quadratic formula, we can calculate the value of y:
y = (-2 ± √(2^2 - 4(1)(7))) / (2(1))
y = (-2 ± √(4 - 28)) / 2
y = (-2 ± √(-24)) / 2

At this point, we encounter an issue because we cannot take the square root of a negative number within the realm of real numbers. Therefore, there is no real solution for y that satisfies the given conditions.

It seems that there might be an error in the problem statement or calculations provided. Please double-check the values for the sides of the triangle or verify if there are any additional constraints or formulas that need to be considered.