First, set the equation equal to zero.

From here, factor the equation into two binomials.

Now, set each binomial equal to zero and solve for x.

 AND 

 AND 

 AND   

Pull out an 

Then factor

So,

To find the roots of this function, we must set it equal to zero and solve for x:

Now, we must find two factors of 12 that add together to get -7. These numbers are -4 and -3. 

We can rewrite this function, then, as two binomials multiplied together:

Now, because this product equals zero, each of the binomials can be set equal to zero:

Our two roots are .

This problem could be worked out using the quadratic formula, but in this particular case it's easier to factor the left side. 

 and  are the roots that zero the expression on the left side of the equation. In the graph, the curve - which happens to be a paraboloa - will cross the x-axis at the roots. 

A few more points...

Observe that the coefficient for the  term in the original quadratic is the sum of   and . Also, the constant term in the originl equation is the product of    and . It's a good rule of thumb to look for numbers that will satsify these conditions when you are setting off to solve a quadratic. Observe how this happens, 

If you notice this pattern in a quadratic, then factoring is always a faster approach. The quadratic formula will always work too, but may take a little longer.

Unfortunately you will often find that factoring is not an option since you will not always be abe to easily find such a pattern for most quadratics, especially if the roots are not whole number integers, or if one or both of the roots are complex numbers. 

Step 1: Rewrite denominator of 2nd term in factored form

Step 2: Multiply each term by the common denominator

Step 3: Simplify

Step 4: Combine like terms, set equal to zero

Step 5: Factor & solve

To find the roots of a quadratic function, we must find the x values where the function is equal to zero. To do this, we must set the function equal to zero:

Now, we factor:

The factoring can be done using a number of methods.

Now, set each binomial equal to zero and solve for x:

To find the roots, or solutions, of this quadratic equation, first factor the equation.

When factored, it's

.

Then, set each of those expressions equal to 0 and solve for x.

Your solutions are

.

Pull out an  term

Two numbers are needed that add to  and multiply to be . Guess and check results in  and .

Each term must be set equal to 0 to find the roots.

The polynomial is degree 4 so there are 4 roots. To make the roots easier to find the expression can be written as

The roots are 

In order to find the roots, factorize the quadratic.

The multiple to the integer 52 are:

The last set can produce the middle term.

Write the binomials.

Setting the equation equal to zero, we have two equations:

Solving for the equations, we will have  as the possible roots.

The answer is:  

Use the quadratic equation to solve for the roots.

Substitute the values of the polynomial  into the equation.

Simplify the quadratic formula.

Since we have a negative discriminant, we will have complex roots even though there are no real roots.

The roots are:  

One of the possible roots given is: