## Homework Statement

I am meant to look for both points when thiobject is 6m above the ground

the equation is

-5x^2+10x+2=6

my question is how do i solve it for specific points in time,
Do i just make the y term 0 and solve from there?

## The Attempt at a Solution

-5x^2 +10x -4

-5(x-1)^2-5=0

I'm guessing $y$ is the vertical displacement and $x$ is some other parameter (the point or points you want to find at which $y = 6$m). Thus you need to solve for $x$ given $y = 6$m.

So all you need to do is solve the equation for $x$ like you have already started doing (either using the quadratic equation or by factorizing like you have done). You have however made a slight mistake in your factoring.

It should be

$-5{(x-1)}^2 + 1=0$

Multiply this out and you should see that this gives you the original expression.

From this it is clear that (subtract 1 and divide by -5 on both sides).

${(x-1)}^2 = \frac{1}{5}$

$x-1 = \pm \sqrt{\frac{1}{5}}$

$x = 1 \pm \sqrt{\frac{1}{5}}$

You could also have used the quadratic equation:

$x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$

To solve the general quadratic: $ax^2 + bx + c = 0$.
Thus in your case we solve for x like so...

$x = \frac{-10 \pm \sqrt{10^2-4(-5)(-4)}}{2(-5)}$

$= \frac{-10 \pm \sqrt{100-80}}{-10}$

$= \frac{-10 \pm \sqrt{20}}{-10}$

$= 1 \mp \sqrt{\frac{20}{100}}$

$= 1 \mp \sqrt{\frac{1}{5}}$

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Thank you for the help, I am just curious as to where you get the +1 constant from.

You have my gratitude. :D

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In case the +1 is still bothering you, consider first the expression that you ended up with.

$-5{(x-1)}^2-5=0$

This expression should be consistent with the original, when one multiplies out to see if it does agree with the initial equation one gets.

$-5x^2 -5(-2x) -5(1) -5= 0$

$-5x^2 +10x -5 -5= 0$

We see now that in order to get -4 and not -10 we need to replace the -5 with a +1.