A differentiable function f(x, y) has the property that f(5,4)=2 and fx(5,4)=−7 and fy(5,4)=1. Find the equation of the tangent plane at the point on the surface z=f(x, y) where x=5, y=4.

"standard form:

3x2 + 14x − 10 = 0

solutions based on quadratic formula:

x1   = −14 − √ 142 − 4×3×(−10) /   2×3   = −14 − 2 × √ 79 /   6   ≈ −5.296

x2   = −14 + √ 142 − 4×3×(−10) /   2×3   = −14 + 2 × √ 79 /   6   ≈ 0.629

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