Solving a quadratic equation without expanding

Question

Solve $(t-1)^2 + 4(t-1) -12 = 0$ without expanding.

My first thought was to solve by expanding $(t-1)^2$ to $(t+1)(t-1)$ and then divide by $(t-1)$, bu obviously that leaves $\frac{-12}{t-1} $. I could of course then complete the square, but I suspect that there is a more elegant solution that I am missing. Any suggestions?

Answer 1

Treat $(t-1)$ as a single variable. Applying the formula, $$t-1=\dfrac{-4\pm\sqrt{4^2-4(-12)}}{2}=-2\pm4=2,-6$$ Then $$t=3,-5$$

Answer 2

Guide:

Let $y=t-1$, solve for $y$ by using factorization, then recover $t=y+1$.

Answer 3

Set $y:=t-1$.

Completing the square:

$(y+2)^2 -4 -12=0;$

$(y+2)^2-16=0.$

Use : $a^2-b^2= (a-b)(a+b)$.

$((y+2)-4)((y+2)+4)=0.$

Hence:

1)$(y+2) -4=0$; or

2)$ (y+2)+4=0.$

Solve for $y$, and recall $y=t-1$.