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-7x^2 + 6x + 3 = 0
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Use \[x = \frac{ (-b \pm \sqrt{b ^{2}-4ac}) }{ 2a }\]
where a = -7, b = 6, c = 3
solve -7x^2 + 6x + 3 = 0 Whats the anwser a.one solution b.two solutions c.no solutions d.infinitely many solutions
In this case, since the \[\pm \sqrt{b ^{2}-4ac}\] is positive, there are two different solutions
what about -8x^2 - 8x - 2 = 0 ?
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Use the same formula but with a = -8, b = -8 and c = -2, if the "\[b ^{2}-4ac\] is greater than zero, there are two solutions, if it is equal to zero there is only one solution, and if it's less than zero there are no (real) solutions
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