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.
<pre>When the DISCRIMINANT (b<sup>2</sup>  -  4ac) > 0, and is a PERFECT SQUARE, then the ROOTS are <b><u>REAL</b></u>, RATIONAL and UNEQUAL.
When the DISCRIMINANT (b<sup>2</sup>  -  4ac) > 0, and is a NON-PERFECT SQUARE, then the ROOTS are <b><u>REAL</b></u>, IRRATIONAL and UNEQUAL.
When the DISCRIMINANT (b<sup>2</sup>  -  4ac) = 0, then the ROOTS are <b><u>REAL</b></u>, RATIONAL and EQUAL.
Therefore, in this case, since you need TWO (2) REAL solutions, then I'd say that, with the DISCRIMINANT {{{b^2  -  4ac >= 0}}}, and the equation, {{{3x^2 + 7x + h = 0}}}, we get:
{{{7^2 - 4(3)(h) >= 0}}}
{{{49 - 12h >= 0}}}
{{{- 12h >= - 49}}}
{{{matrix(1,3, h <= (- 49)/(- 12), "======>", h <= 49/12)}}}