Question and answer
Solve this quadratic equation. x2 + 5x + 3 = 0
Question
Asked 4/29/2011 2:12:14 PM
Updated 7/17/2014 4:56:59 AM
2 Answers/Comments
This conversation has been flagged as incorrect.
Flagged by andrewpallarca [7/17/2014 4:53:09 AM]
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Original conversation
User: What is the solution set of 7x2 + 3x = 0?

User: Solve this quadratic equation. x2 + 5x + 3 = 0

Weegy: x=(-5-sqrt(13))/2 and x=(sqrt(13)-5)/2
mchlklck|Points 2172|

User: What is the solution set of (x - 2)(x - 3) = 2?

Weegy: Do you have choices? I think the answer is -2/7
ihcvin|Points 185|

User: What is the solution set of (x - 2)(x - 3) = 2? {1, 4} {2, 3} {4, 5}

Weegy: y+4=2x
fireandh20|Points 1550|

User: What is the solution set of (x - 2)(x - 3) = 2?

User: The square of a certain negative number is equal to five more than one-half of that number. What is the number?

User: The width and the length of a rectangle are consecutive even integers. If the width is decreased by 3 inches, then the area of the resulting rectangle is 24 square inches. What is the area of the original rectangle?

Question
Asked 4/29/2011 2:12:14 PM
Updated 7/17/2014 4:56:59 AM
2 Answers/Comments
This conversation has been flagged as incorrect.
Flagged by andrewpallarca [7/17/2014 4:53:09 AM]
New answers
Rating
8
7x^2 + 3x = 0;
x(7x + 3) = 0;
x = 0 or 7x + 3 = 0; 7x = -3; x = -3/7

Solution set is {0, -3/7}
Added 7/17/2014 4:53:07 AM
This answer has been confirmed as correct and helpful.
8
(x - 2)(x - 3) = 2;
x^2 - 5x + 6 = 2;
x^2 - 5x + 6 - 2 = 0;
x^2 - 5x + 4 = 0;
(x - 1)(x - 4) = 0;
(x - 1) = 0; x = 1 or (x - 4) = 0; x = 4

the solution set is (x = 1, x = 4)
Added 7/17/2014 4:56:59 AM
This answer has been confirmed as correct and helpful.
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