In order to solve this problem, we need to combine like terms and then factor:

$\displaystyle (x-3)^2=2x^2+14$

$\displaystyle x^2-6x+9=2x^2+14$

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

$\displaystyle 0=(x+5)(x+1)$

$\displaystyle x=-1, -5$

To solve this we must factor:

The first step is to recognize that all of the terms on the equation's left side contain 4x. Therefore we can pull 4x out:

$\displaystyle 4x(x^{2}-7x +12) = 0$

Then we factor the inside of the parentheses, realizing that only -3, and -4 add to form -7, and multiply to form 12:

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

Now we can see that, for the left side of the equation to equal zero, x can only equal 0, 3, or 4.

To solve, divide out a $\displaystyle 2$ and then factor.

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

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

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

$\displaystyle x=4,-1$

$\displaystyle \frac{3x^{2}-27}{x-3}=\frac{3(x^{2}-9)}{x-3}=\frac{3(x-3)(x+3)}{x-3}$

$\displaystyle =3(x+3)=3x+9.$

$\displaystyle \left | 3x - 7 \right |=8$ really consists of two equations: $\displaystyle 3x - 7 = \pm 8$

We must solve them both to find two possible solutions.

$\displaystyle 3x - 7 = 8 \Rightarrow 3x = 15\Rightarrow x = 5$

$\displaystyle 3x - 7 = - 8 \Rightarrow 3x = -1\Rightarrow x = -1/3$

So $\displaystyle x=5$ or $\displaystyle x=$ $\displaystyle \dpi{100} -\frac{1}{3}$.

It's actually easier to solve for the complement first.  Let's solve $\displaystyle \left | 2x-5 \right |< 3$.  That gives -3 < 2x - 5 < 3.  Add 5 to get 2 < 2x < 8, and divide by 2 to get 1 < x < 4.  To find the real solution then, we take the opposites of the two inequality signs.  Then our answer becomes $\displaystyle x\leq 1 \textsc{ or } x\geq 4$.

Set $\displaystyle f (x) = 0$ and solve for $\displaystyle x$:

$\displaystyle f (x) = 0$

$\displaystyle \left | 4x + B \right | - 8 = 0$

$\displaystyle \left | 4x + B \right | = 8$

Rewrite as a compound equation and solve each part separately:

$\displaystyle 4x + B = -8 \textrm{ or } 4x + B = 8$

$\displaystyle 4x + B = -8$

$\displaystyle 4x + B -B = -B -8$

$\displaystyle 4x = -B -8$

$\displaystyle 4x \div 4 =\left ( -B -8 \right ) \div 4$

$\displaystyle x = \frac{-B -8 }{4}$

$\displaystyle 4x + B = 8$

$\displaystyle 4x + B -B = -B + 8$

$\displaystyle 4x = -B +8$

$\displaystyle 4x \div 4 =\left ( -B +8 \right ) \div 4$

$\displaystyle x = \frac{-B +8 }{4}$

We can rewrite this as an equation, where $\displaystyle N$ is the number in question:

$\displaystyle N = |N| - 10$

A nonnegative number is equal to its own absolute value, so if this number exists, it must be negative.

In thsi case, $\displaystyle |N| = - N$, and we can rewrite that equation as

$\displaystyle N = -N- 10$

$\displaystyle N +N= -N- 10 +N$

$\displaystyle 2N = -10$

$\displaystyle 2N \div 2 = -10\div 2$

$\displaystyle N = -5$

This is the only number that fits the criterion.

Since $\displaystyle -1< n< 0$, we know the following:  

$\displaystyle 0< n^2< 1$ ;

$\displaystyle -1< \frac{n}{2}< 0$;

$\displaystyle \frac{1}{n^2}>1$;

$\displaystyle \frac{1}{n}< -1$;

$\displaystyle 0< n+1< 1$.

Also, we need to compare absolute values, so the greatest one must be either $\displaystyle \left | \frac{1}{n^2}\right |$ or $\displaystyle \left | \frac{1}{n}\right |$.

We also know that $\displaystyle \left | n^2\right |< \left | n\right |$ when $\displaystyle -1< n< 0$.

Thus, we know for sure that $\displaystyle \left | \frac{1}{n^2}\right |>\left | \frac{1}{n}\right |$.

We can rewrite this as an equation, where $\displaystyle N$ is the number in question:

$\displaystyle N = 2\cdot |N| - 20$

If $\displaystyle N$ is nonnegative, then $\displaystyle | N | = N$, and we can rewrite this as 

$\displaystyle N = 2N - 20$

Solve:

$\displaystyle N -2N= 2N - 20-2N$

$\displaystyle -N= - 20$

$\displaystyle N = 20$

If $\displaystyle N$ is negative, then $\displaystyle | N | =- N$, and we can rewrite this as 

$\displaystyle N = 2\left (-N \right ) - 20$

$\displaystyle N = -2N - 20$

$\displaystyle N + 2N= -2N - 20 + 2N$

$\displaystyle 3N=- 20$

$\displaystyle 3N \div 3=- 20\div 3$

$\displaystyle N=- 6\frac{2}{3}$

The numbers $\displaystyle - 6\frac{2}{3}, 20$ have the given characteristics.