First multiply both sides by \(\displaystyle 3x\):

\(\displaystyle \frac{2}{3}-\frac{3}{x}=\frac{8}{x}\Rightarrow 3x(\frac{2}{3})-3x(\frac{3}{x})=3x(\frac{8}{x})\Rightarrow 6x-9=24\Rightarrow 6x=33\)

\(\displaystyle \Rightarrow x=5.5\)

\(\displaystyle 2\left |x+5 \right |-5< 7\Rightarrow 2\left |x+5 \right |-5+5< 7+5\Rightarrow 2\left |x+5 \right |< 12\Rightarrow \left |x+5 \right |< 6\)

\(\displaystyle \left |x \right |< a\Rightarrow -a< x< a\)

We have \(\displaystyle \left |x+5 \right |< 6\), so:

\(\displaystyle -6< x+5< 6\Rightarrow -6-5< x+5-5< 6-5\Rightarrow -11< x< 1\)

\(\displaystyle \left |2x+4 \right |-7>9\Rightarrow \left |2x+4 \right |-7+7>9+7\Rightarrow \left |2x+4 \right |>16\)

\(\displaystyle \left |x \right |>a\)  means that  \(\displaystyle x< -a\)   or   \(\displaystyle x>a\). 

Then we can write:

\(\displaystyle \left |2x+4 \right |>16\Rightarrow 2x+4>16\) 

\(\displaystyle 2x+4< -16\)

\(\displaystyle 2x+4>16\Rightarrow 2x+4-4>16-4\Rightarrow 2x>12\Rightarrow x>6\)

\(\displaystyle 2x+4< -16\Rightarrow 2x+4-4< -16-4\Rightarrow 2x< -20\Rightarrow x< -10\)

\(\displaystyle 7\sqrt{x}+11=5\Rightarrow 7\sqrt{x}+11-11=5-11\Rightarrow 7\sqrt{x}=-6\Rightarrow \sqrt{x}=-\frac{6}{7}\)

Since \(\displaystyle \sqrt{x}\) cannot be negative, no real number \(\displaystyle x\) satisfies the equation.

First simplify the equation by dividing both sides by \(\displaystyle 3\):

\(\displaystyle 3x^2-6x-45=0\Rightarrow x^2-2x-15=0\)

If \(\displaystyle a, b, c\) are real numbers with \(\displaystyle a\neq 0\), and if \(\displaystyle ax^2+bx+c=0\), then 

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

The expression \(\displaystyle b^2-4ac\) is called the discriminant of the quadratic equation, and we say \(\displaystyle D=b^2-4ac\) .

We can write \(\displaystyle x=\frac{-b\pm \sqrt{D}}{2a}\).

In this problem we have \(\displaystyle a=1,\ b=-2, \ and\ c=-15.\)

\(\displaystyle D=b^2-4ac=(-2)^2-4\times 1\times (-15)=4+60=64\)

\(\displaystyle x=\frac{-b\pm \sqrt{D}}{2a}=\frac{-(-2)\pm \sqrt{64}}{2\times 1}=\frac{2\pm 8}{2}\)

Therefore the roots are

\(\displaystyle x=\frac{2+8}{2}=5\)    or    \(\displaystyle x=\frac{2-8}{2}=-3\).

First rewrite the equation in the form of \(\displaystyle ax^2+bx+c=0\):

\(\displaystyle x^2+3x-28=0\)

\(\displaystyle 4x^2-64=0\Rightarrow 4x^2=64\Rightarrow x^2=16\Rightarrow x=\pm \sqrt{16}=\pm 4\)

Therefore, \(\displaystyle x=4\)  or  \(\displaystyle x=-4\).

We can factor out an \(\displaystyle x\):

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

Therefore, \(\displaystyle x=0\) or:

 \(\displaystyle 4x-3=0\Rightarrow 4x-3+3=0+3\Rightarrow 4x=3\Rightarrow x=\frac{3}{4}\)

Sometimes you can easily factor the experssion \(\displaystyle ax^2+bx+c=0\). Then the equation can be solved by setting each factor equal to \(\displaystyle 0\). In this problem we have:

\(\displaystyle x^2-x-20=0\Rightarrow (x-5)(x+4)=0\Rightarrow x-5=0\)  or  \(\displaystyle x+4=0\).

\(\displaystyle x-5=0\Rightarrow x=5\)

\(\displaystyle x+4=0 \Rightarrow x=-4\)