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For what value(s) of b will cause 6x^2+bx-7=0 to have two real number solutions

User Vvill
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1 Answer

11 votes

Answer:

B can be All real values.

Explanation:

The discriminant = b^2 - 4*6*-7 > 0

b^2 + 168 > 0

b^2 > -168

All values of b will be > 0.

User Jeremiah Polo
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