- Rewrite the equation in standard form: \(10x^2 - 6x - 3 = 0\).
- Identify the coefficients: \(a=10, b=-6, c=-3\).
- Substitute into the quadratic formula and calculate the discriminant: \(b^2 - 4ac = 156\).
- Simplify the roots: \(\boxed{\frac{3+\sqrt{39}}{10}, \frac{3-\sqrt{39}}{10}}\).
Explanation
- Writing in Standard Form
To solve the equation \(10x^2 = 6x + 3\) using the quadratic formula, we first need to express it in the standard quadratic form \(ax^2 + bx + c = 0\). By subtracting \(6x\) and \(3\) from both sides, we obtain:
\[10x^2 - 6x - 3 = 0.\]
Here, we can identify the coefficients as:
\(a = 10\)
\(b = -6\)
\(c = -3\)
- Calculating the Discriminant
The quadratic formula is given by:
\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}.\]
We calculate the discriminant, which is the part inside the square root, \(D = b^2 - 4ac\):
\[D = (-6)^2 - 4(10)(-3)\]
\[D = 36 + 120\]
\[D = 156.\]
- Solving and Simplifying
Now we substitute the values of \(a, b,\) and the discriminant into the quadratic formula:
\[x = \frac{-(-6) \pm \sqrt{156}}{2(10)}\]
\[x = \frac{6 \pm \sqrt{156}}{20}.\]
Next, we simplify the square root \(\sqrt{156}\). Since \(156 = 4 \times 39\), we have \(\sqrt{156} = \sqrt{4 \times 39} = 2\sqrt{39}\).
Substituting this back into our expression for \(x\):
\[x = \frac{6 \pm 2\sqrt{39}}{20}.\]
We can simplify this fraction by dividing the numerator and denominator by 2:
\[x = \frac{3 \pm \sqrt{39}}{10}.\]
Thus, the solutions are \(x = \frac{3 + \sqrt{39}}{10}\) and \(x = \frac{3 - \sqrt{39}}{10}\).
Examples
The quadratic formula is fundamental in physics and engineering for calculating the trajectories of objects or finding critical values in circuits. For example, when launching a ball, its height over time is defined by a quadratic function; finding when the ball hits the ground involves solving a quadratic equation for time. Understanding how to use the quadratic formula ensures you can model these real-world scenarios accurately and predict outcomes reliably.