Q:

How do I find the solution(s) to my second-degree equation? 

Solve the second-degree equation $x^2 - 3x=4$

My attempt : 
$$
\begin{align}
x^2-3x &= 4\\
x(x-3) &= 4\\
x-3 &= 4 \\
x &= 7\\
\end{align}
$$
I managed to solve one part of this problem but that one part is wrong.  I don't understand how to solve for $x$, when $x(x-3) = 4$. I can solve for the $x$ inside the bracket but I don't know how to solve the one outside the bracket and the $x$ I solved for inside the bracket is somehow wrong. 

A:

$$x(x-3)=4$$
$$x^2-3x-4=0$$
$$(x-4)(x+1)=0$$
Are you able to complete the rest? 
Extra comment: In your working, from $x(x-3)=4$, we cannot conclude that $(x-3)=4$.
Extra comment 2: note that $$(x-a)(x-b)=x^2-(a+b)x+ab$$
Hence to recover $a$ and $b$, I tried for factor of $-4$ where the sum is equal to $3$. 
For general quadractic equation, you can complete the square or use the quadractic formula.