I have a question for using quadratic equation again. I have 3^2-5x-7=0 Can any one help me?

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I have a question for using quadratic equation again. I have 3^2-5x-7=0 Can any one help me?

Mathematics
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Do you mean 3x^2?
is nt in the form of quadartic
yes I meant 3x^2-5x-7=0

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Other answers:

(3x+7)(x-1)
no sorry
x[1] = 1/6 (5-sqrt(109)) x[2] = 1/6 (5+sqrt(109))
OK doesn't factorise so use formula: x =( 5 +/- sqrt(25 - 4 times 3 times -7))/3 I get x = (5 +/- sqrt109)/3
?
Do you know the quadratic formula? The solution of ax^2 + bx + c = 0 is \[x = -b +/- \sqrt{(b^2 - 4ac}) \over 2a\]
yes I know that
@ suzi20 would it not be this x=-5+sqrt109/6, x=-5-sqrt109/6
OK so look at suzi20's correct solution and see if you can get the same answer by substituting for a, b and c in the formula, where a = 3, b = -5 and c = -7
leahp x=-5+sqrt109/6, x=-5-sqrt109/6 is a correct answer
thank you, again
I have one more question does the letter i is not also use for the symbol +- sign together

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