SOLUTION: Find all values of h for which the quadratic equation has two real solutions. 3x^2+7x +h=0. Write your answer as an equality or inequality in terms of h.

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Question 1144405: Find all values of h for which the quadratic equation has two real solutions. 3x^2+7x +h=0. Write your answer as an equality or inequality in terms of h.
Found 4 solutions by greenestamps, ikleyn, Alan3354, MathTherapy:
Answer by greenestamps(13198)   (Show Source):
You can put this solution on YOUR website!

A quadratic equation ax^2+bx+c=0 has real solutions if b^2-4ac is non-negative.

For your example,

Answer by ikleyn(52777)   (Show Source):
You can put this solution on YOUR website!
.

A quadratic equation ax^2 + bx + c = 0 has two distinct real roots if and only if the discriminant b^2-4ac is POSITIVE. For your example, > 49-12h > 0 49 > 12h h < 49/12

Compare my solution and my answer with that of the tutor @greenestamps and notice
that my inequality signs are always STRICT inequalities to provide real solutions, as the problem requires.

If you admit merging two solutions into one, then you can use the solution by @greenestamps.

If you do not admit such merging, then you should use my solution and my answer.

            Unfortunately, the exact meaning of the problem and of the question is not clear from the post.

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To learn everything about quadratic equations, look into the lessons
    - Introduction into Quadratic Equations
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    - HOW TO solve quadratic equation by completing the square - Learning by examples
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Answer by Alan3354(69443)   (Show Source):
You can put this solution on YOUR website!
Quadratics always have 2 solutions.
If they are real, sometimes the 2 solutions are the same (when the Disc = 0).
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You should specify 2 different real solutions if that's what you mean.

Answer by MathTherapy(10551)   (Show Source):
You can put this solution on YOUR website!
Find all values of h for which the quadratic equation has two real solutions. 3x^2+7x +h=0. Write your answer as an equality or inequality in terms of h.

When the DISCRIMINANT (b² - 4ac) > 0, and is a PERFECT SQUARE, then the ROOTS are REAL, RATIONAL and UNEQUAL.
When the DISCRIMINANT (b² - 4ac) > 0, and is a NON-PERFECT SQUARE, then the ROOTS are REAL, IRRATIONAL and UNEQUAL.
When the DISCRIMINANT (b² - 4ac) = 0, then the ROOTS are REAL, RATIONAL and EQUAL.
Therefore, in this case, since you need TWO (2) REAL solutions, then I'd say that, with the DISCRIMINANT , and the equation, , we get: