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# Algebra 1

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I don't know how to solve x^2-2x-3, could someone help with the steps, but don't give me the answer?

Guest Sep 5, 2017
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#1
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I mean x^2-2x-3=0

Guest Sep 5, 2017
#2
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$$x = {-b \pm \sqrt{b^2-4ac} \over 2a}$$

In the equation: x2-2x-3

a=1

b=-2

c=-3

Guest Sep 5, 2017
#3
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x2 - 2x - 3  =  0

You can solve for  x  with the quadratic formula as Guest said, or you can factor the expression on the left side. To factor it.....you need two numbers that add to  -2  and multiply to  -3 .

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Here's an example that might help without giving away the answer.

x2 - 6x + 8  =  0

What two numbers add to  -6  and multiply to 8  ?  →      -4  and  -2

So, we can factor the left side like this....

(x - 4)(x - 2)  =  0

Now set each factor equal to zero and solve for  x  .

x - 4  =  0                              x - 2  =  0

x  =  4                 or               x  =  2

So  x = 4  causes this expression to equal zero, and also  x = 2  causes the expression to equal zero.

hectictar  Sep 5, 2017
edited by hectictar  Sep 5, 2017

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