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Write the equation in the form ax² + bx + c = 0, then enter a, b, and c. The value of a cannot be zero, since that would make the equation linear.
The roots are x = (-b ± square root of (b² - 4ac)) / (2a). The value b² - 4ac is the discriminant: positive gives two real roots, zero gives one repeated real root, and negative gives two complex roots.
For x² - 5x + 6 = 0, enter a = 1, b = -5, and c = 6. The formula gives x = 2 and x = 3.
The equation has no real roots. It has two complex roots containing the imaginary unit i.
No. If a is zero, the x² term disappears and the equation should be solved as a linear equation.
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