Question 1207406
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Note it is possible, or even probable, that you have not asked the second question correctly.<br>
As implied in the statement of the problem, every quadratic equation has two unequal real solutions, a repeated real solution, or no real solutions.  There is no quadratic equation that has "no solution at all".<br>
The given equation has no real solutions: b^2-4ac = 16-28 = -12 < 0.<br>
By definition, any perfect square quadratic polynomial has a repeated real solution (or, better stated, two equal solutions): x^2-6x+9 --> b^2-4ac = 36-36 = 0.<br>
For an example with two real solutions, start with any quadratic polynomial that is a factor of two linear polynomials.  Example: (x+3)(x-2) = x^2+x-6 --> b^2-4ac = 1+24 = 25 > 0.<br>