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# Solving Second Degree Polynomials

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A second degree polynomial, also referred as a quadratic equation can be expressed as below:
ax2 + bx + c = 0

to solve the equation we can use the quadratic formulas as shown below:

x1 = (-b + (b2-4ac)1/2)/2a
x2 = (-b - (b2-4ac)1/2)/2a

a quadratic equation has two solutions when b2-4ac > 0

a quadratic equation has only one solution when b2-4ac = 0

a quadratic equation has no solution when b2-4ac < 0

Example (2 solutions)
2x2 + 6x + 1 = 0

b2-4ac = 62-4 x 2 x 1 = 28, since 28 > 0, we can conclude that there exists two solutions

x1 = (-b + (b2-4ac)1/2)/2a = -0.177
x2 = (-b - (b2-4ac)1/2)/2a = -2.822

Example (1 solutions)
3x2 + 6x + 3 = 0

b2-4ac = 62-4 x 3 x 3 = 0, thus we can conclude that there only exists one solution

x = -b/2a = -1

How to construct a quadratic equation when its solutions are given
if x1 = 3 and x2 = 2, then we can construct the equation as shown below:

p(x) = (x - x1)(x - x2) = (x - 3)(x - 2) = x2 - 5x + 6 = 0.