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14-02-2017
Mathematics

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use the quadratic formula to solve 2w²+w=5

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Аноним Аноним

14-02-2017

2w²+w=5
⇒2w²+w-5 = 0
here a=2, b=1 and c = -5

[tex]x = \frac{-b \pm \ (b^2-4ac) }{2a} \\ \\ \Rightarrow x = \frac{-1 \pm \ \sqrt{(1)^2-4*2*(-5)} }{2*2} \\ \\ \Rightarrow x = \frac{-1 \pm \ \sqrt{1+40} }{2*2} \\ \\ \Rightarrow x= \frac{-1+ \sqrt{41} }{4} \ and\ \frac{-1- \sqrt{41} }{4} [/tex]

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