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

2 Answers
Mar 18, 2018

#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))#

Mar 18, 2018

#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#