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The quadratic formula is a powerful tool that helps solve any quadratic equation of the form \( ax^2 + bx + c = 0 \). The formula is expressed as:\[y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]This formula gives solutions to quadratic equations by providing the values of \( y \) that make the equation true. Each component of the formula has specific roles:

• \( -b \) changes the sign of \( b \).
• \( \pm \sqrt{b^2 - 4ac} \) encompasses the square root of the discriminant and indicates two potential values for \( y \): one summing the square root, the other subtracting it.
• \( 2a \) is the denominator that scales down the equation.

In our exercise, the values \( a = 2 \), \( b = -1 \), and \( c = -\frac{1}{2} \) are substituted into the quadratic formula to solve for the real roots of the equation.

The discriminant is a key part of the quadratic formula found inside the square root: \( b^2 - 4ac \). It plays an essential role in determining the nature of the roots of a quadratic equation:

• If the discriminant is positive, the quadratic equation has two distinct real solutions.
• If it's zero, there is exactly one real solution, often called a double root.
• If the discriminant is negative, the equation has no real solutions.

In our specific example, the discriminant is calculated as \( \Delta = 1^2 - 4(2)(-\frac{1}{2}) = 5 \). Since 5 is a positive number, we conclude that the equation possesses two distinct real solutions. Calculating the discriminant first also provides insight into the complexity of solving the equation.

Real solutions are the values that satisfy the quadratic equation, placing it in a true statement when substituted back into the original equation. For the equation \( 2y^2 - y - \frac{1}{2} = 0 \), we employ the quadratic formula and find that the real solutions are:\[y = \frac{1 + \sqrt{5}}{4} \quad \text{and} \quad y = \frac{1 - \sqrt{5}}{4}\]These real solutions are derived when the discriminant is positive, leading to two different values for \( y \). Thus, when you plug these values for \( y \) back into the original equation, it should hold true for both cases. Real solutions give clarity that the equation's graph touches the x-axis at distinct points, signifying where real roots exist. Understanding real solutions helps in interpreting results accurately and ensures a firm grasp of quadratic equations.