How do you solve #y^2+y=20#?

Answer 1

Kennedy Adams

#y^2 +y = 20#

#y^2 + y - 20 =0#

What #2# numbers add to #+1# and multiply to #-20#?

#+5# and #-4#.

Therefore,

#(y+5)(y-4) = 20#

Therefore, #y# can be #color(red)(-5 or, +4#

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Answer 2

Kennedy Adams

#y = -5, 4#

#y^2+y=20#.

Subtracting #20# From Both Sides,

#y^2+y-20 = 0#.

Factoring,

#(y+5)(y-4) = 0#.

Since One Of The Answers Have To Be 0,

#y+5 = 0#, #y = -5#

#y-4 = 0#, #y = 4#

The Answers Are: #-5# and #4#.

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Answer 3

Jace Anderson

To solve the equation (y^2 + y = 20), you can first rearrange it into a quadratic equation in standard form:

(y^2 + y - 20 = 0)

Then, you can use factoring, completing the square, or the quadratic formula to find the solutions for (y). In this case, factoring might be the simplest approach:

(y^2 + 5y - 4y - 20 = 0)

(y(y + 5) - 4(y + 5) = 0)

((y - 4)(y + 5) = 0)

So, the solutions for (y) are (y = 4) and (y = -5).

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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.