How do you solve #6r^2+5r=1# using the quadratic formula?
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To solve the equation 6r^2 + 5r = 1 using the quadratic formula, follow these steps:

Identify the coefficients a, b, and c in the quadratic equation in the form ax^2 + bx + c. a = 6 b = 5 c = 1

Substitute the values of a, b, and c into the quadratic formula: r = (b ± √(b^2  4ac)) / (2a)

Plug in the values: r = ((5) ± √((5)^2  4(6)(1))) / (2(6))

Simplify the expression inside the square root: = ((5) ± √(25 + 24)) / 12 = ((5) ± √(49)) / 12

Since the square root of 49 is 7: r = ((5) ± 7) / 12

Solve for both possible values of r: r₁ = (5 + 7) / 12 = 2/12 = 1/6 r₂ = (5  7) / 12 = 12/12 = 1
Therefore, the solutions to the equation 6r^2 + 5r = 1 are r = 1/6 and r = 1.
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When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
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