The quadratic formula comes from the standard form of a quadratic equation.In this form, we have three important coefficients: a, b, and c.These coefficients are the key to using the quadratic formula.The quadratic formula itself may look complicated, but we can break it down into manageable parts.Let's examine each part of the formula.Let's look at a specific example: x squared plus five x plus six equals zero.In this equation, we can identify our coefficients.The coefficient of x squared is one, the coefficient of x is five, and the constant term is six.These values will be substituted into our formula in their corresponding positions.Now that we have our values, let's plug them into the quadratic formula.We substitute a equals 1, b equals 5, and c equals 6 into the formula.Let's simplify what's under the square root. Five squared is twenty-five, and four times one times six is twenty-four.Twenty-five minus twenty-four equals one under the square root.The square root of one is simply one.Now let's solve for both the plus and minus cases separately.For the plus case, negative five plus one equals negative four, divided by two gives us negative two.For the minus case, negative five minus one equals negative six, divided by two gives us negative three.Therefore, our equation has two solutions: x equals negative two and x equals negative three.In the next section, we'll verify these solutions by graphing the parabola.Now let's verify our solutions graphically by plotting the parabola.Here's our quadratic function y equals x squared plus five x plus six.Our calculated solutions were x equals negative two and x equals negative three. Let's mark these points on the x-axis.Let's verify that x equals negative two is a solution by plugging it back into our equation.Similarly, let's verify that x equals negative three is also a solution.These points where the parabola crosses the x-axis, called x-intercepts, represent the solutions to our quadratic equation.When we plug these x-values back into the original equation, we get y equals zero, confirming they are indeed our solutions.
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