How to find $x$ from expression? 
Solve for $x$ in the expression $$x(2x + 5) = 168.$$

I have already tried to move $x$ from one side to another with brackets and without them, like this:
$x = 168;\quad 2x + 5 = 168 \implies 2x = 163 \implies x = 81.5$
but eventually without success. 
I know that the answer is $x = 8$ but I can't come to this solution.
How to solve this?
 A: if $f(x)g(x) = A$ that does not mean that $f(x)=A$ or $g(x) = A$. However, what you could do is expand as follows
$$2x^2 + 5x = 168$$
then
$$2x^2 + 5x - 168 = 0$$
then you have to realize that what you have is a quadratic equation, where normally you compute a quantity called the discriminant
$$\Delta = b^2 - 4ac$$
for a quadratic equation $ax^2 + bx + c $. In your case, you got $a = 2,b=5,c=-168$ so
$$\Delta = (5)^2 - 4(2)(-168) = 1369$$
Now since $\Delta > 0$, you have two distinct solutions,
$$x_1 = \frac{-b - \sqrt{\Delta}}{2a}=\frac{-5 - \sqrt{1369}}{4}$$
$$x_2 = \frac{-b + \sqrt{\Delta}}{2a}=\frac{-5 + \sqrt{1369}}{4}$$
A: That's a quadratic equation ($ax^2+bx+c=0$) because $x$ there occurs to a second power. Multiply each term in the parentheses by $x$ and move $168$ over to the left side:
$$
x(2x + 5) = 168\implies\\
2x^2 + 5x - 168=0.
$$
Now, use the quadratic formula to find the roots (the solutions to your equation):
$$
x_{1,2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}.
$$
Your $a$ there is $2$, $b$ is $5$ and $c$ is $-168$. You will see that $-10.5$ is another solution to your equation.
A: Using middle term factor
$$0=2x^2+5x-168=2x^2+(21-16)x-168=x(\underbrace{2x+21})-8(\underbrace{2x+21})=?$$
A: Write
$$2x^2+5x-168=0$$
or
$$16x^2+40x-1344=0$$
and complete the square:
$$16x^2+40x-1344=16x^2+2\cdot5\cdot4x+5^2-5^2-1344=(4x+5)^2-37^2=0.$$
Under this form, the resolution is easy.

Note that there is total cheating in the answer as you can't guess upfront that multiplying by $8$ leads to integer expressions. In the normal way, you would divide by $2$ and work with fractions. Anyway, the final answer is the same.
