# How do you solve the quadratic using the quadratic given 2n^2+6n-56?

Apr 17, 2018

n=-7, 4

#### Explanation:

You can solve this using factoring
$2 {n}^{2} + 6 n - 56$
$2 \left({n}^{2} + 3 n - 28\right)$
$2 \left(n + 7\right) \left(n - 4\right)$
When $n$ is equivalent to -7 or 4 , then the equation is equal to 0, and that is how you solve it.

Apr 17, 2018

x=4 and x=7 by dividing and factoring, factoring, or using the quadratic equation.

#### Explanation:

Method 1, divide and factor.
In this method, you try to get the x^2 term equal to 1x^2. This can be done in this expression by dividing by two because every term is evenly divisible by two. This gives you the expression ${x}^{2} + 3 x - 28$
Your next step is to factor. You are looking for factors of -28 that add up to 3. In this case, it is -4 and 7. You can then set the next expression up. It is (x-4)(x+7) Assuming that you are solving for zero, you can use the zero products property, which says that anything times zero is zero. This means you can set the expression x-4=0 and x+7=0, yielding x=4 and x=-7.

Method 2, factor.
You are pretty much doing what is above, but without dividing by two first. Instead of looking for factors of-28 that add up to 3, you are looking for factors fo -112, or -56*2 that add up to 6. In this case, it is -8 and 14. The next step is to set the expression up. In this case, it is $\left(2 {x}^{2} + 14 x\right) + \left(- 8 x - 56\right)$ If you will notice, this is still the same equation as you had originally, but the 6x term is broken up. After you have done the setup, you get to factor again. The goal is to factor out as much as you can from each expression. From the first one, you can take out 2x, giving you 2x(x+7). The second one can be divided by -8, yielding (x+7). Again, if you will notice, after factoring out as much as you can, the two expressions have an equal term, the (x+7). If you don't see something like that on other problems, check your work because something is wrong. The next step is to set up the final expression. Take what you factored out and combine them, then multiply it by one of the extra ones. This gives you (2x-8)(x+7). Using the zero products property, solve as above.

The quadratic formula lets you solve any quadratic equation, regardless of whether or not it can be factored. It is $\frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$, when your quadratic is in the form ax^2+bx+c=0. Assuming that is what you are solving for, your a term is 2, your b term is 6, and your c term is -56. Putting it through the formula yields -7 and 4 as your answers.