I have:
$$T(x) = -\frac1{10}x^2 + \frac{24}{10}x - \frac{44}{10}$$
$T$ = temperature
$x$ = hour
And they ask me to find the hour in which the temperature reached zero. So I know I must replace:
$$0 = -\frac1{10}x^2 + \frac{24}{10}x - \frac{44}{10}$$
It's very easy, I know, but I really don't remember how to clear/free the $x$ in this case.
On
Let's multiply by $-10$ to clear the fractions and get $$0=x^2-24x+44.$$ It follows, then, that $$\begin{align}100 &= x^2-24x+144\\ &=x^2+2(-12)x+(-12)^2\\ &= (x-12)^2.\end{align}$$ The two numbers whose square is $100$ are $\pm 10$, so we have $x-12=\pm 10$, from which we have $x=2$ or $x=22$.
On
You want to solve the quadratic equation:
$$ 0 = -\frac{1}{10}x^2 + \frac{24}{10}x - \frac{44}{10} $$ You can indeed just apply the quadratic formula, but first you can make things a tiny bit easier by multiplying by -10 on both sides to get $$ 0 = x^2 - 24x + 44. $$ Now if $a$ and $b$ are the solutions you can also factor the polynomial as $$ (x-a)(x-b) = x^2 - (a+b) + ab. $$ So if you wanted to try and guess the solutions, you are trying to find two numbers $a$ and $b$ such that $-(a + b) = -24$ and $ab = 44$. Maybe it is not too hard too see that $a = 22$ and $b = 2$. So in all you get the two solutions to the original equation: $$x = 22 \quad \text{and} \quad x = 2. $$
Use the quadratic formula so that the zeroes of the function occur at $\frac{-b \pm \sqrt {b^{2}-4ac}}{2}$, where $a=\frac{-1}{10},b=\frac{24}{10}x,c=\frac{-44}{10}$. In general, polynomials of up to degree 4 have solutions in roots, but solutions of degree 5 or higher do not.