SOLUTION: 2x^2+3xy-4y^2 when x=2 and y = -4

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Question 430280: 2x^2+3xy-4y^2 when x=2 and y = -4
Answer by algebrahouse.com(1659) About Me  (Show Source):
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2x² + 3xy - 4y² when x = 2 and y = -4
= 2(2)² + 3(2)(-4) - 4(-4)² {substituted values in}
= 2(4) - 24 - 4(16) {evaluated exponents and multiplied}
= 8 - 24 - 64 {multiplied}
= -80 {combined like terms}
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