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Solve the equation for all real solutions in simplest form.5, r, squared, minus, 7, r, plus, 4, equals, 65r 2 −7r+4=6

Question

Solve the equation for all real solutions in simplest form.

Given the equation:

5r27r+4=6 5r^2 - 7r + 4 = 6

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Solution

To solve the equation 65r^2 - 7r + 4 = 6, we need to rearrange the equation to bring all terms to one side and set it equal to zero.

First, subtract 6 from both sides of the equation: 65r^2 - 7r + 4 - 6 = 0 65r^2 - 7r - 2 = 0

Next, we can use the quadratic formula to find the values of r that satisfy the equation. The quadratic formula states that for an equation of the form ax^2 + bx + c = 0, the solutions are given by:

x = (-b ± √(b^2 - 4ac)) / (2a)

In our equation, a = 65, b = -7, and c = -2. Plugging these values into the quadratic formula, we get:

r = (-(-7) ± √((-7)^2 - 4(65)(-2))) / (2(65)) r = (7 ± √(49 + 520)) / 130 r = (7 ± √569) / 130

Therefore, the solutions to the equation 65r^2 - 7r + 4 = 6 are: r = (7 + √569) / 130 r = (7 - √569) / 130

These are the real solutions in simplest form.

This problem has been solved

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