In algebra, a quadratic (from the Latin quadrates for "square") is an equation that will be rearranged in standard form as

ax2 + bx + c = 0

Just put the values of a, b and c into the Quadratic Formula, and do the calculations.
If a=0, the equation is not quadratic.

#### Example: Solve $$5x^2 + 6x + c = 0$$

When $$a \ne 0$$, there are two solutions to $$ax^2 + bx + c = 0$$ and they are $x = {-b \pm \sqrt{b^2-4ac} \over 2a}.$

Coefficients are: $$a=5 . b=6 . c = 1$$
Quadratic Formula: Put in a, b and c:

$x = {-6 \pm \sqrt{(6^2-4×5×1)} \over 2×5}.$

$x = {-6 \pm \sqrt{(36-20)} \over 10}.$

$x = {-6 \pm \sqrt{(16)} \over 10}.$

$x = {-6 \pm (4) \over 10}.$

$x = {-6 \pm 4 \over 10}.$

$x = -0.2 OR -1.$

# What is the Quadrate equation?

In algebra, a quadratic (from the Latin quadrates for "square") is an equation that will be rearranged in standard form as

ax2 + bx + c = 0

Where x represents an unknown, and a, b, and c represent known numbers, where a ≠ 0. If a = 0, then the equation is linear, not quadratic, as there's no ax2 term. The numbers a, b, and c are the coefficients of the equation and should be distinguished by calling them, respectively, the quadratic coefficient, the linear coefficient, and therefore the constant or free term.

If you'll rewrite your equation during this form, it means it is often solved with the quadratic formula. an answer to the present equation is additionally called a root of an equation.

The quadratic formula is as follows:    x = (-B ± √Δ)/2A

Where:  Δ = B² - 4AC

Using this formula, you'll find the solutions to any quadratic. Note that there are three possible options for obtaining a result:

The quadratic has two unique roots when Δ > 0. Then, the primary solution of the quadratic formula is xâ‚ = (-B + √Δ)/2A, and therefore the second is xâ‚‚ = (-B - √Δ)/2A.

The quadratic has just one root when Δ = 0. The answer is adequate to x = -B/2A. It’s sometimes called a repeated or double root.

The quadratic has no real solutions for Δ < 0.

You can also graph the function y = Ax² + Bx + C. Its shape may be a parabola, and therefore the roots of the quadratic are the x-intercepts of this function.

## How to use the quadratic formula to solve the equation

Write down your equation. Let's assume it's 4x² + 3x - 7 = -4 - x.

Bring the equation to the shape Ax² + Bx + C = 0. During this example, we'll within the roll in the hay in the following steps:

4x² + 3x - 7 = -4 - x

4x² + (3+1)x + (-7+4) = 0

4x² + 4x - 3 = 0

Calculate the determinant.

Δ = B² - 4AC = 4² - 4*4*(-3) = 16 + 48 = 64.

Decide whether the determinant is bigger, equal, or less than 0. In our case, the determinant is bigger than 0, which suggests that this equation has two unique roots.

Calculate the 2 roots using the quadratic formula.

xâ‚ = (-B + √Δ)/2A = (-4 +√64) / (2*4) = (-4+8) / 8 = 4/8 = 0.5

xâ‚‚ = (-B - √Δ)/2A= (-4 -√64) / (2*4) = (-4-8) / 8 = -12/8 = -1.5

The roots of your equation are xâ‚ = 0.5 and xâ‚‚ = -1.5.

You can also simply type the values of A, B, and C into our quadratic calculator and let it perform all calculations for you.

Make sure you've got written down the right number of digits using our significant figures calculator.

### How to use this Quadrate equation

This tool is really great and it is a really free and web-based tool. You can use this tool to solve a lot of problems. All students and any individual can use this tool. This tool can be used all over the world.

Now, let’s understand how to use this tool.

As we know the formula, you have to enter the value properly that you have given the question.

After you will enter all the values in the box, you have to click on the calculate button to get the result.

This is how this tool works.

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