How do you solve the inequality x^2-6x-7<0

Explanation:
First, factorise the x terms.
${\displaystyle x^2-6x-7<0}$
${\displaystyle (x-7)(x+1)<0}$
To find the roots:
${\displaystyle x-7=0}$ or ${\displaystyle x+1=0}$
${\displaystyle x=7}$ or ${\displaystyle x=-1}$
The question asks us to find the portion of the curve which is less than zero ${\displaystyle \left(<0\right)}$. Thus, we need to show by giving a range of values of x.
Do a rough sketch of a graph showing
${\displaystyle x^2}$ curve on x and y axis
- curve must be positive, meaning it has a minimum point. (like a smile)
- label the x-intercepts with the values found above.
Since the equality given is "<", draw a small circle at the intercepts.
The portion of the curve that is negative falls below ${\displaystyle y=0}$
This means the curve is negative between ${\displaystyle x=-1}$ and ${\displaystyle x=7}$
Which is expressed as ${\displaystyle -1<x<7}$
(If the question states ${\displaystyle x^2-6x-7>0}$, the portions of the curve that are positive is when ${\displaystyle x<-1}$ and ${\displaystyle x>7}$)