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If the discriminant of the quadratic equation is equal or greater than zero it will have 2 solutions if it is less than zero then there are no solutions.

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Q: How do you find quadratic equation with only 2 solutions?
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How you find the solution of a quadratic equation by graphing its quadratic equation?

When you graph the quadratic equation, you have three possibilities... 1. The graph touches x-axis once. Then that quadratic equation only has one solution and you find it by finding the x-intercept. 2. The graph touches x-axis twice. Then that quadratic equation has two solutions and you also find it by finding the x-intercept 3. The graph doesn't touch the x-axis at all. Then that quadratic equation has no solutions. If you really want to find the solutions, you'll have to go to imaginary solutions, where the solutions include negative square roots.


How do you know if a quadratic equation will have one two or no solutions How do you find a quadratic equation if you are only given the solution Is it possible to have different quadratic equation?

Draw the graph of the equation. the solution is/are the points where the line cuts the x(horisontal) axis .


Is it possible for a quadratic equation to have more than 2 solutions?

No because quadratic equations only have 2 X-Intercepts


How do you find a quadratic equation with only the solution?

If the solutions are p and q, then the quadratic is (x-p)(x-q) = 0 or x2 - (p+q)x + pq = 0 Hope this is what the question meant!


Could you ever have three solutions to a quadratic equation?

No. By definition, a quadratic equation can have at most two solutions. For a quadratic of the form ax^2 + bx + c, when the discriminant of a quadratic, b^2 - 4a*c is positive you have two distinct real solutions. As the discriminant becomes smaller, the two solutions move closer together. When the discriminant becomes zero, the two solutions coincide which may also be considered a quadratic with only one solution. When the discriminant is negative, there are no real solutions but there will be two complex solutions - that is those involving i = sqrt(-1).