A free quadratic equation calculator that shows and explains each step in solving your quadratic equation.

You entered:

There are no solutions in the real number domain.

There are two complex solutions: x = 0.3 + 0.60377599699347

where

(1)

For any quadratic equation

(2)

In the form above, you specified values for the variables a, b, and c. Plugging those values into Eqn. 1, we get:

(3) \(x=--33\pm\frac{\sqrt{-33^2-4*55*25}}{2*55}\)

which simplifies to:

(4) \(x=--33\pm\frac{\sqrt{1089-5500}}{110}\)

(5) \(x=--33\pm\frac{\sqrt{-4411}}{110}\)

This means that our solution will require finding the square root of a negative number. There is no real number solution for this, so our solution will be a complex number (that is, it will involve the imaginary number

Let's calculate the square root:

(6) \(x=--33\pm\frac{66.415359669281i}{110}\)

This equation further simplifies to:

(7) \(x=-\frac{--33}{110}\pm0.60377599699347i\)

Solving for x, we find two solutions which are both complex numbers:

x = 0.3 + 0.60377599699347

and

x = 0.3 - 0.60377599699347

Both of these solutions are complex numbers.

These are the two solutions that will satisfy the equation

Calculating a solution to a quadratic equation may appear daunting, because both x and x

Solving a quadratic equation will always result in 2 solutions for x. These solutions are called roots. These roots may both be real numbers or, they may both be complex numbers. Rarely, both roots may be the same, producing one solution for x.

Why do we need to be able to solve quadratic equations? Quadratic equations are needed to compute answers to many real-world problems. For example, to calculate how an object will rise and fall due to Earth's gravity would require the use of s quadratic equation.

The term "quadratic" comes from the Latin word

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