How do you solve #5x ^ { 2} + 9x - 20= 4- 5x#?

Answer 1

Rearrange the expression so one side equals zero, then use the quadratic formula to find #x={-4,1.2}#

First, we'll rearrange the expression so the Right Hand Side (RHS) equals zero. This will give us our quadratic expression to use in the quadratic formula:

#5x^2+9x-20color(red)(-4)color(blue)(+5x)=cancel(4)color(red)(cancel(-4))cancel(-5x)color(blue)(cancel(+5x))#
#5x^2+9xcolor(blue)(+5x)-20color(red)(-4)=0#
#5x^2+14x-24=0#
Now, we'll utilize The Quadratic Formula. For any quadratic expression that looks like #ax^2+bx+c#:
#x=(-b+-sqrt(b^2-4ac))/(2a)#

For this case:

#a=5# #b=14# #c=-24#

Let's plug that in:

#x=(-14+-sqrt((14^2)-4(5)(-24)))/(2(5))#
#x=(-14+-sqrt(196-4(-120)))/10#
#x=(-14+-sqrt(196+480))/10#
#x=(-14+-sqrt(676))/10#
#x=(-14+-26)/10#
#x={(-14-26)/10,(-14+26)/10}#
#x={(-40)/10,12/10}#
#color(green)(x={-4,1.2})#
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Answer 2

To solve (5x^2 + 9x - 20 = 4 - 5x):

  1. Bring all terms to one side to set the equation to zero.
  2. Rearrange terms to have the quadratic equation in standard form.
  3. Use the quadratic formula or factorization to solve for (x).
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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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