Notice in this question there are only two terms, the exponent value and the constant value. There is also the negative sign between the two. When we look at each number we see that each are a perfect square. Due to the negative sign between the two, this type of quadratic expression can also be written as a difference of squares. We look at the exponential term and see it is

The perfect square factors of this term are   and  .

Now we look at the constant term

The perfect square factors of this term are   and 

Now to combine these into the binomial factor form we need to remember it is the difference of perfect squares meaning we will have one subtraction sign and one adding sign, so we get the following:

From here we solve each binomial for x. To do this we set each binomal to zero and solve for x.

Notice in this question there are only two terms, the exponent value and the constant value. There is also the negative sign between the two. When we look at each number we see that each are a perfect square. Due to the negative sign between the two, this type of quadratic expression can also be written as a difference of squares. We look at the exponential term and see it is

The perfect square factors of this term are  and  .

Now we look at the constant term

The perfect square factors of this term are   and  

Now to combine these into the binomial factor form we need to remember it is the difference of perfect squares meaning we will have one subtraction sign and one adding sign, so we get the following:

From here we solve each binomial for x. To do this we set each binomal to zero and solve for x.

Notice in this question there are only two terms, the exponent value and the constant value. There is also the negative sign between the two. When we look at each number we see that each are a perfect square. Due to the negative sign between the two, this type of quadratic expression can also be written as a difference of squares. We look at the exponential term and see it is

The perfect square factors of this term are  and .

Now we look at the constant term

The perfect square factors of this term are  and . 

Now to combine these into the binomial factor form we need to remember it is the difference of perfect squares meaning we will have one subtraction sign and one adding sign, so we get the following:

From here we solve each binomial for x. To do this we set each binomal to zero and solve for x.

Notice in this question there are only two terms, the exponent value and the constant value. There is also the negative sign between the two. When we look at each term we see that each is a perfect square. Due to the negative sign between the two, this type of quadratic expression can also be written as a difference of squares. We look at the exponential term and see it is

The perfect square factors of this term are  and .

Now we look at the constant term

The perfecct square factors of this term are  and .

Now to combine these into the binomial factor form we need to remember it is the difference of perfect squares meaning we will have one subtraction sign and one adding sign, so we get the following:

From here we solve each binomial for x. To do this we set each binomal to zero and solve for x.

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. 

To solve by completing the square, you should first take the numerical coefficient to the “right side” of the equation:

Then, divide the middle coefficient by 2:

Square that and add it to both sides:

Now, you can easily factor the quadratic:

Your next step would be to take the square root of both sides. At this point, however, you know that you cannot solve the problem. When you take the square root of both sides, you will be forced to take the square root of . This is impossible (at least in terms of real numbers), meaning that this problem must have no real solution.

If  is one odd number, then the next odd number is . If their product is 143, then the following equation is true.

Distribute into the parenthesis.

Subtract 143 from both sides.

This can be solved by factoring, or by the quadratic equation. We will use the latter.

We are told that both integers are positive, so .

The other integer is .

Start by changing the less than sign to an equal sign and solve for .

Now, plot these two numbers on a number line.

Notice how the number line is divided into three regions:

Now, choose a number fromeach of these regions to plug back into the inequality to test if the inequality holds.

For , let 

Since this number is not less than zero, the solution cannot be found in this region.

For , let 

Since this number is less than zero, the solution can be found in this region.

For  let .

Since this number is not less than zero, the solution cannot be found in this region.

Because the solution is only negative in the interval , that must be the solution.

First, we can factor the quadratic to give us a better understanding of its graph. Factoring gives us: . Now we know that the quadratic has zeros at  and . Furthermore this information reveals that the quadratic is positive. Using this information, we can sketch a graph like this: 

We can see that the parabola is below the x-axis (in other words, less than ) between these two zeros  and .

The only x-value satisfying the inequality  is .

The value of  would work if the inequality were inclusive, but since it is strictly less than instead of less than or equal to , that value will not work.