m2-5m-14=0

b2-4b+4=0

2m2+2m-12=0

8n2+7n-15=7

and the hardest!

k2-31-2k=-6-3k2-k

Some points to remember:

1. Be sure to set your equations equal to 0.

2. See if the equation factors easily.

3. If it does, factor, then set each factor equal to 0.

4. Solve for the solutions.

Let's look at your first equation:

m^2 - 5m - 14 = 0
(I'm using ^2 to mean squared.)

This equation is already equal to 0.

Let's see if it factors easily:
(m - 7)(m + 2) = 0 (It does!)

Now let's set each factor equal to 0:
m - 7 = 0; m = 7
m + 2 = 0; m = -2

Check the solutions with the original equation by substituting the values you found for m. It always helps to check your work.

I hope this will help get you started.

m2-5m-14=0

To solve each of these equations, you will need to use algebraic techniques such as factoring, completing the square, or using the quadratic formula. Let's solve each equation step by step:

1) m^2 - 5m - 14 = 0

To solve this quadratic equation, we can factor it. We need to find two numbers whose product is -14 and whose sum is -5. The numbers are -7 and 2. So, factoring the equation:

(m - 7)(m + 2) = 0

Setting each factor equal to zero:

m - 7 = 0 or m + 2 = 0

Solving for m in each equation gives us two possible solutions:

m = 7 or m = -2

2) b^2 - 4b + 4 = 0

This equation is a perfect square trinomial, since b^2 and 4 are perfect squares. It can be factored as:

(b - 2)^2 = 0

Taking the square root of both sides:

b - 2 = 0

Solving for b gives us:

b = 2

3) 2m^2 + 2m - 12 = 0

This equation can be factored by grouping. Factor out a common factor from the first two terms and the last two terms:

2m^2 + 2m - 12 = 2(m^2 + m) - 12 = 2m(m + 1) - 12

Now, we have a quadratic trinomial that can be factored:

2m(m + 1) - 12 = (2m - 6)(m + 1) = 0

Setting each factor equal to zero:

2m - 6 = 0 or m + 1 = 0

Solving for m gives us:

m = 3 or m = -1

4) 8n^2 + 7n - 15 = 7

To solve this equation, we need to rearrange it to be an equal to zero:

8n^2 + 7n - 22 = 0

To factor this quadratic equation, we look for two numbers whose product is -22 and whose sum is 7. The numbers are 11 and -2. Factoring:

(8n - 2)(n + 11) = 0

Setting each factor equal to zero:

8n - 2 = 0 or n + 11 = 0

Solving for n gives us:

n = 1/4 or n = -11

5) k^2 - 31 - 2k = -6 - 3k^2 - k

To solve this equation, we need to simplify and rearrange it:

k^2 - 31 - 2k = -6 - 3k^2 - k

Combining like terms:

k^2 + 3k^2 - 2k + k = -6 + 31

Simplifying further:

4k^2 - k - 25 = 0

This is a quadratic equation. To solve it, you can use factoring, completing the square, or the quadratic formula. Factoring is not straightforward for this equation, so let's use the quadratic formula:

The quadratic formula states that for an equation of the form ax^2 + bx + c = 0, the solutions are given by:

x = (-b ± √(b^2 - 4ac)) / 2a

For our equation, a = 4, b = -1, and c = -25. Substituting these values into the quadratic formula:

k = (-(-1) ± √((-1)^2 - 4(4)(-25))) / (2(4))

Simplifying:

k = (1 ± √(1 + 400)) / 8
k = (1 ± √401) / 8

So the solutions for this equation are:

k = (1 + √401) / 8 and k = (1 - √401) / 8

These are the solutions for the given equations.