Question

How do you solve `-2t ( t + 2) = - 3`?

Answer 1

`t=(4+-sqrt40)/(-4)`

Explanation:

`-2t(t+2)=-3`

Use the distributive property for the left side

`(-2t)(t)+(-2t)(2)=-3`

`-2t^2 - 4t =-3`

Then subtract `color(red)(-3)` on both sides

`-2t^2 - 4t - color(red)(-3)=-3 - color(red)(-3)`

`-2t^2 - 4t + 3 = -3 + 3`

`-2t^2 - 4t + 3 = 0`

Now we can use the quadratic formula with `a=-2, b=-4, c=3`

`t= ((-b)+-sqrt(b^2-4ac))/(2a)`

`t=(-(-4)+-sqrt((-4)^2-4(-2)(3)))/((2)(-2))`

`t=(4+-sqrt(16+24))/(-4)`

`=>color(green)(t=(4+-sqrt(40))/(-4))`

An equivalent simplified version:

`t=(4+-2sqrt(10))/(-4)`

`t=-1+-(-1/2sqrt(10))`

`=>t=-1+-(-sqrt(5/2))`

Answer 2

`t~~-2.581`
`t~~0.581`

Explanation:

You need to multiply the `-2t` with the parentheses and make the equation equal to 0.
`-2t(t+2)=-3`
`-> -2t^2 -4t=-3`
`->-2t^2-4t+3=0`

You now have a quadratic equation.

For the rest of this answer, I took a lot of this page (I encourage you to go take a look at it)

I will use the quadratic formula to solve it.

`t=(-b+-sqrt(b^2-4ac))/(2a)`

where `at^2+bt+c=0`

For `-2t^2-4t+3=0` :

`a=-2`
`b=-4`
`c=3`

Now substitute into the quadratic formula:
`t=(-(-4)+-sqrt((-4)^2-4(-2)(3)))/(2(-2))`
`t=(4+-sqrt(16+24))/(-4)`
`t=(4+-sqrt(40))/(-4)`

Therefore:
`t=(4+sqrt(40))/(-4)-> ~~-2.581`
OR
`t=(4-sqrt(40))/(-4)-> ~~0.581`