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3 years ago

How would you find the solutions to this quadratic equation y=2x^2-6x-8

Mathematics

2 answers:

aleksandrvk [35]3 years ago

8 0

Answer:

Factoring.

General Formulas and Concepts:

Algebra I

Factoring
Finding roots of quadratics

Step-by-step explanation:

Step 1: Define

y = 2x² - 6x - 8

0 = 2x² - 6x - 8

Step 2: Find roots

Factor 2: 0 = 2(x² - 3x - 4)
Factor: 0 = 2(x - 4)(x + 1)
Find roots: x = -1, 4

And we have our final answer!

Marysya12 [62]3 years ago

7 0

Answer:

first take 2 out of the equation

y=2(x^2-3x-4)

then factor

y=2(x-4)(x+1)

then solve

(x-4)=0 . (x+1)=0

x=4 . x=-1

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Take the second equation and flip it around so the y on the left ends up on the right and the 4x on the right ends up on the left. This makes all negatives positive and all positives negative. -4x + 12 = y

Then add the first equation to the second equation
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So if you know that y = negative 4, you can substitute into either equation. I pick the second one because I am a lazy person.
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So your solution is: x = 4 and y = -4 or this is also written (4, -4)

Does that work for you?

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