SOLUTION: need intercepting points for these two equations: f(x)=-2/315x^2+630 and 4/3x+420 thnx!

Question 927990: need intercepting points for these two equations: f(x)=-2/315x^2+630 and 4/3x+420
thnx!
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
need intercepting points for these two equations:
f(x)=-2/315x^2+630 and 4/3x+420
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Solve::
(-2/315)x^2 + 630 = (4/3)x + 420
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(-2/315)x^2 - (4/3)x + 210 = 0
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I graphed it and found these Real Number solutions::
x = 105 and x = -315
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When x = 105, y = (-2/315)105^2 + 630 = 560
When x = -315, y = (-2/315)315^2 + 630 = 0
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Cheers,
Stan H.
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