Question 107339
Given:
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3x^2+5x-3=x <=== OK
3x^2+5x-3-x=0 <=== OK
3x^2+4x=3 <=== OK, but why not solve for x by using the quadratic formula
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Start with your second equation:
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3x^2 + 5x - 3 -x = 0
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Combine the 5x and -x to get +4x. This reduces the equation to:
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3x^2 + 4x - 3 = 0
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Now recognize that this is in the standard quadratic form of ax^2 + bx + c = 0 in which (by 
comparing your equation with the standard quadratic form) a = 3, b = 4, and c = -3. 
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The quadratic formula says that when you have a quadratic equation in the standard quadratic
form, the answer for x is given by:
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{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}
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So to get the answer for x all you have to do is appropriately substitute 3 for a, 4 for b, 
and -3 for c in the answer equation. When you do that you get:
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{{{x = (-(4)+-sqrt(4^2 - 4*3*(-3)))/(2*3)}}}
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notice that -(4) is just -4. Then within the radical sign 4^2 is 16 and -4*3*(-3) is +36.
And finally in the denominator 2*3 = 6. Substitute these values into the equation and you get:
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{{{x = (-4+-sqrt(16 +36))/(6)}}}
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The terms under the radical sign add up to 52. And this 52 factors to 4*13. Substituting this
into the answer for x results in:
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{{{x = (-4+-sqrt(4*13))/(6)}}}
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But {{{sqrt(4*13) = sqrt(4)*sqrt(13) = 2*sqrt(13)}}}
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Substituting this for sqrt(4*13) further simplifies the problem to:
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{{{x = (-4+-2*sqrt(13))/(6)}}}
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Notice that both terms in the numerator have 2 as a factor. And the denominator also has
2 as a factor. So, if you divide the numerator by 2 and the denominator by 2, the equation
for x reduces to:
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{{{x = (-2+-sqrt(13))/(3)}}}
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Now you can write the two answers for x. First use the plus sign between the terms in the 
numerator to get:
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{{{x = (-2+sqrt(13))/(3)}}}
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then use the minus sign to get the other answer for x:
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{{{x = (-2-sqrt(13))/(3)}}}
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And there you have the two answers for x
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Hope this helps you understand the problem. 
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