# Quadratic Formula

Applying the quadratic formula is the most practical and widely used method to solve quadratic equations. Quadratic equations are polynomial equations of the second degree, meaning that the highest power of the variable is two. The quadratic formula is a formula that provides the solution to any quadratic equation in the standard form and it is widely used because it is simple to understand, easy to apply, and provides accurate solutions for any quadratic equation.

Given a quadratic equation in the standard form \(ax^2+bx+c=0\), the quadratic formula is:

\[ x_{1,2} = \frac{{-b \pm \sqrt{{b^2-4ac}}}}{{2a}} \]

- \(a, b, c\) are the coefficients of the equations and \(a \neq 0\).
- The plus-minus symbol indicates that the quadratic equation has two solutions.
- A quadratic equation always has at most two solutions if a real double root is counted for two \((x = x_1, x = x_2)\), and complex roots are included.

It’s important to note that the quadratic formula only works for quadratic equations, which are equations of the form \(ax^2+bx+c = 0\), where \(a, b\) and \(c\) are constants and \(a\) is not equal to zero.

## Discriminant

The term within the square root, \(\Delta = b^2 – 4ac\), is known as the *discriminant*, and it is crucial in determining the quadratic equation’s nature and number of solutions.

If \( \Delta > 0\), the quadratic equation has two distinct real solutions. \[S = \{x_1, x_2\} \quad x_1, x_2 \in \mathbb{R} \quad x_1 \neq x_2 \] \[ x_{1,2} = \frac{{-b \pm \sqrt{{b^2-4ac}}}}{{2a}}\]

If \( \Delta = 0\), the quadratic equation has two coincident real solutions. \[S = \{x\} \quad x \in \mathbb{R} \quad x = x_1 = x_2 \] \[x=\frac{-b}{2a}\]

If \( \Delta = < 0\), the quadratic equation has no real solutions. Instead, it gives rise to complex solutions characterized by imaginary components. \[\nexists \hspace{10px} x \in \mathbb{R}\] \[x_{1,2} = \frac{-b \pm i \sqrt{|b^2-4ac|}}{2a}\]

In a second-degree equation with a non-negative discriminant, the sum of the roots is given by the ratio \(\frac{-b}{a}\) and their product by \( \frac{c}{a} \).

## Example

Solve the equation \(\; x^2-4x+2=0\)

In order to solve a quadratic equation using the quadratic formula, we must first write the equation in standard form \(ax^2 +bx+c = 0\). After that, we can substitute the values of \(a\), \(b\), and \(c\) into the quadratic formula. So, if the equation is already in standard form, we can simply substitute its coefficients into the quadratic formula.

\[x_{1,2} = \frac{{-b \pm \sqrt{{b^2-4ac}}}}{{2a}}\]

We obtain:

\[x_{1,2} = \frac{{-(-4) \pm \sqrt{{(-4)^2-4(1)(+2)}}}}{{2a}} \]

In this case, the discriminant is \(\Delta = 9-8 \geq 0\), so the equation admits two distinct real solutions.

\begin{align*} x_{1,2} &= \frac{{4 \pm \sqrt{{16-8}}}}{2} \\[0.8em] &= \frac{{4 \pm 1 }}{2} \\[0.6em] \end{align*}

Finally, by performing the calculations, we obtain:

\[ x_1 = -\frac{3}{2} \] \[ x_2 = \frac{5}{2} \]

The solution to the equation is: \[ x_1 = -\frac{3}{2} \quad x_2 = \frac{5}{2} \]