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use the quadratic formula to find the zeros in the function. y-6=-2x^2
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i don't get what you're saying @dalia_lam
Have you considered factoring? \(2x^{2} + y - 6 = 0\) Go!
How about solving y = -2x° + 6
Oh, sorry. It says to use the quadratic formula. Do it! a = 2 b = 1 c = -6
a = -2 b = 0 c = 6 x=-b +-sqrt(b²-4ac) / 2a x = +-sqrt(-4*-2*6) / 2*-2 tkhunny - wouldn't b=0 and a =-2
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continuing ......... x = sqrt(48) / -4 = 2 * sqrt(12) / -4 x = -sqrt (12)
thanks you so much!
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x=4*sqrt(3)/-4 x = -sqrt(3) and x = sqrt(3)
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