How do you make "-5(t^2)+10t+35" into factored form?
First you take out a common factor of $\displaystyle -5$ leaving
$\displaystyle -5t^2+10t+35 = -5(t^2-2t-7)$
Now I would complete the square inside the brackets.
$\displaystyle -5(t^2-2t-7)$
You take the coeffeicent of the linear term, half it then square it.
$\displaystyle \left( \frac{-2}{2}\right)^2 = (-1)^2 = 1$ giving
$\displaystyle -5((t^2-2t{\color{red}+1})-7{\color{red}-1})$
$\displaystyle -5((t-1)^2-8)$
$\displaystyle -5((t-1)^2-(\sqrt{8})^2)$
Now applying the difference of 2 squares and we are finished.
$\displaystyle -5(t-1-\sqrt{8})(t-1+\sqrt{8})$
In $\displaystyle -5(t^2-2t-7)$ I was looking for a new constant term that would help me make a perfect square. Finidng $\displaystyle +1$ as this term gave
$\displaystyle -5((\underbrace{t^2-2t{\color{red}+1}}_{\text{perfect square}})-7{\color{red}-1})$
I couldn't just throw away the $\displaystyle -7$ and to balance the equation I introduced an extra $\displaystyle -1$ from the $\displaystyle +1$ found.
A perfect square has the from of either
$\displaystyle a^2-2ab+b^2 = (a-b)^2$ or $\displaystyle a^2+2ab+b^2 = (a+b)^2$