First, set the equation equal to zero.

$\displaystyle 0=-8x^2-14x-3$

From here, factor the equation into two binomials.

$\displaystyle 0=(-2x-3)(4x-1)$

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

$\displaystyle 0=-2x-3$ AND $\displaystyle 0=4x-1$

$\displaystyle 3=-2x$ AND $\displaystyle 1=4x$

$\displaystyle \frac{3}{-2}=x$ AND   $\displaystyle \frac{1}{4}=x$

Pull out an $\displaystyle x^2$

$\displaystyle x^2(x^2+2x+1)=0$

Then factor

$\displaystyle x^2(x+1)(x+1)=0$

So,

 $\displaystyle x^2=0;(x+1)=0$

$\displaystyle x=0; x=-1$

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

$\displaystyle x^2-7x+12=0$

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:

$\displaystyle (x-4)(x-3)=0$

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

$\displaystyle x-4=0$

$\displaystyle x=4$

$\displaystyle x-3=0$

$\displaystyle x=3$

Our two roots are $\displaystyle x=3, 4$.

$\displaystyle x^2 -x -6 = 0$

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

$\displaystyle (x+2)(x-3)=0$

$\displaystyle x = \left{-2 \right }$ and $\displaystyle x = 3$ 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 $\displaystyle \large x$ term in the original quadratic is the sum of  $\displaystyle 2$ and $\displaystyle -3$. Also, the constant term in the originl equation is the product of   $\displaystyle 2$ and $\displaystyle -3$. 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, 

$\displaystyle \large (x+a)(x+b)$

$\displaystyle \large = x^2 +ax +bx + ab$

$\displaystyle \large = x^2 + (a+b)x +ab$

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

$\displaystyle 6+\frac{12}{(x+1)(x-1)}=\frac{5}{x-1}$

Step 2: Multiply each term by the common denominator

$\displaystyle 6*(x+1)(x-1)+\frac{12*(x+1)*(x-1)}{(x+1)(x-1)}=\frac{5*(x+1)(x-1)}{x-1}$

Step 3: Simplify

$\displaystyle 6x^2-6+12=5x+5$

Step 4: Combine like terms, set equal to zero

$\displaystyle 6x^2-5x+1=0$

Step 5: Factor & solve

$\displaystyle (3x-1)(2x-1)=0$

$\displaystyle x=\frac{1}{2},\frac{1}{3}$

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:

$\displaystyle 4x^2-3x-1=0$

Now, we factor:

$\displaystyle (4x+1)(x-1)=0$

The factoring can be done using a number of methods.

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

$\displaystyle x=-\frac{1}{4}, 1$

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

When factored, it's

$\displaystyle (x-7)(x+3)$.

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

Your solutions are

$\displaystyle x=-3,7$.

$\displaystyle x^4-6x^3+9x^2$

Pull out an $\displaystyle x^2$ term

$\displaystyle x^2(x^2-6x+9)$

Two numbers are needed that add to $\displaystyle -6$ and multiply to be $\displaystyle 9$. Guess and check results in $\displaystyle -3$ and $\displaystyle -3$.

$\displaystyle x^2(x-3)(x-3)$

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

$\displaystyle xx(x-3)(x-3)$

The roots are $\displaystyle 0,0,3,3$

In order to find the roots, factorize the quadratic.

The multiple to the integer 52 are:

$\displaystyle [1,52] , [2,26], [4,13]$

The last set can produce the middle term.

Write the binomials.

$\displaystyle y=x^2-17x +52 = (x-4)(x-13)$

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

$\displaystyle x-4=0$

$\displaystyle x-13=0$

Solving for the equations, we will have $\displaystyle x=4,13$ as the possible roots.

The answer is:  $\displaystyle 4,13$

Use the quadratic equation to solve for the roots.

$\displaystyle x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

Substitute the values of the polynomial $\displaystyle y=ax^2+bx+c$ into the equation.

$\displaystyle y=2x^2-3x+20$

$\displaystyle x=\frac{-(-3)\pm\sqrt{(-3)^2-4(2)(20)}}{2(2)}$

Simplify the quadratic formula.

$\displaystyle x=\frac{+3\pm\sqrt{9-160}}{4} = \frac{3\pm\sqrt{-151}}{4}$

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

The roots are:  $\displaystyle x=\frac{3\pm i\sqrt{151}}{4}$

One of the possible roots given is:  $\displaystyle \frac{3- i\sqrt{151}}{4}$