### Instruction: The height of a ball when it is thrown off a cliff can be represented by the equation $h=45-7t-6t^2$, where $t$ is time in seconds. In how many seconds will the ball reach a height of 25 feet?### Response: I need to find the value of $t$ that makes $h=25$.
I can do this by setting the equation equal to 25 and solving for $t$.
I get $25=45-7t-6t^2$.
I want to simplify this equation and make it quadratic, so I can use the quadratic formula.
I can do this by subtracting 25 from both sides and rearranging the terms.
I get $6t^2+7t-20=0$.
Now I can apply the quadratic formula to find the values of $t$ that satisfy this equation.
The quadratic formula says that if $ax^2+bx+c=0$, then $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$.
In this case, $a=6$, $b=7$, and $c=-20$.
So I can plug these values into the formula and simplify.
I get $t=\frac{-7\pm\sqrt{7^2-4(6)(-20)}}{2(6)}$.
I get $t=\frac{-7\pm\sqrt{529}}{12}$.
I get $t=\frac{-7\pm 23}{12}$.
I get two possible values for $t$: $t=\frac{16}{12}$ and $t=\frac{-30}{12}$.
However, the value of $t=\frac{-30}{12}$ does not make sense in the context of the problem, because it means the ball reached 25 feet before it was even thrown.
So I can reject that solution and focus on the other one.
I get $t=\frac{16}{12}=\frac{4}{3}$.
This means the ball will reach a height of 25 feet after $\frac{4}{3}$ seconds, or 1.33 seconds.