The quadratic equation is:
\[ ax^2 + bx + c = 0 \]
To solve for \( x \), we use the quadratic formula:
\[ x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}} \]
In algebra, the quadratic formula is the solution to the quadratic equation. The quadratic equation is a second-order polynomial equation in a single variable \( x \) with a non-zero coefficient for \( x^2 \). The quadratic formula provides the solution(s) to the equation:
\[ ax^2 + bx + c = 0 \]
where \( a \), \( b \), and \( c \) are coefficients of the equation, and \( a \neq 0 \).
Let's solve the quadratic equation \( 2x^2 + 4x - 6 = 0 \).
Here, \( a = 2 \), \( b = 4 \), and \( c = -6 \). Using the quadratic formula:
\[ x = \frac{{-4 \pm \sqrt{{4^2 - 4 \cdot 2 \cdot (-6)}}}}{{2 \cdot 2}} \]
Simplifying inside the square root:
\[ x = \frac{{-4 \pm \sqrt{{16 + 48}}}}{{4}} = \frac{{-4 \pm \sqrt{{64}}}}{{4}} \]
Simplifying further:
\[ x = \frac{{-4 \pm 8}}{{4}} \]
So, we have two solutions:
\[ x_1 = \frac{{-4 + 8}}{{4}} = 1 \quad \text{and} \quad x_2 = \frac{{-4 - 8}}{{4}} = -3 \]
Therefore, the roots of the equation \( 2x^2 + 4x - 6 = 0 \) are \( x_1 = 1 \) and \( x_2 = -3 \).
Let's solve the quadratic equation \( x^2 + 2x + 5 = 0 \).
Here, \( a = 1 \), \( b = 2 \), and \( c = 5 \). Using the quadratic formula:
\[ x = \frac{{-2 \pm \sqrt{{2^2 - 4 \cdot 1 \cdot 5}}}}{{2 \cdot 1}} \]
Simplifying inside the square root:
\[ x = \frac{{-2 \pm \sqrt{{4 - 20}}}}{{2}} = \frac{{-2 \pm \sqrt{{-16}}}}{{2}} \]
Since we have a negative number under the square root, we get complex roots:
\[ x = \frac{{-2 \pm 4i}}{{2}} = -1 \pm 2i \]
Therefore, the roots of the equation \( x^2 + 2x + 5 = 0 \) are \( x_1 = -1 + 2i \) and \( x_2 = -1 - 2i \).
