### 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 want to find the value of $t$ that makes $h=25$, so I will substitute 25 for $h$ in the equation and then solve for $t$.
I get $25=45-7t-6t^2$, and I want to rearrange this into a quadratic equation of the form $at^2+bt+c=0$, where $a$, $b$, and $c$ are constants.
To do that, I will subtract 25 from both sides and then simplify.
I get $-6t^2-7t+20=0$.
Now I need to find the roots of this equation, that is, the values of $t$ that make it zero.
There are different methods to do this, such as factoring, completing the square, or using the quadratic formula.
I will use the quadratic formula, which says that if $at^2+bt+c=0$, then $t=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$.
In this case, $a=-6$, $b=-7$, and $c=20$.
So I plug these values into the formula and simplify.
I get $t=\frac{-(-7)\pm\sqrt{(-7)^2-4(-6)(20)}}{2(-6)}$.
I simplify the expression under the square root first.
I get $49+480=529$.
This is a perfect square, so I can take its square root.
I get $\sqrt{529}=23$.
Then I simplify the rest of the expression.
I get $t=\frac{7\pm 23}{-12}$.
This gives me two possible values for $t$.
One is $t=\frac{7+23}{-12}=-\frac{5}{2}=-2.5$, and the other is $t=\frac{7-23}{-12}=\frac{4}{3}\approx 1.33$.
However, I need to check if these values make sense in the context of the problem.
Since $t$ is time in seconds, it cannot be negative.
So I reject the value $t=-2.5$ as extraneous.
The only valid value for $t$ is $t=\frac{4}{3}$.
This means that the ball will reach a height of 25 feet in $\frac{4}{3}$ seconds, or about 1.33 seconds.