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4x^2-3x-5=0
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Solve for x
\[\left\{x\to \frac{1}{8} \left(3-\sqrt{89}\right)\right\},\left\{x\to \frac{1}{8} \left(3+\sqrt{89}\right)\right\} \]
Can you help me understand
http://www.scientificpages.net/math/images/Quadratic_Formula.gif you use the quadratic formula!
I bet robtobey used mathematica use the quadratic formula\[x = (-b \pm \sqrt{b^2-4ac})/(2a) \] a =4 b = -3 c = -5 plug n play
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u could also use the fact that; \[x1*x2 = -b/a\] and \[x1+x2 = c/a\]
Ok got it thanks
Quick question why does the 8 become 1/8 ?
Thanks
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