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Solve the given quadratic equation using the quadratic formula. 6t^(2)+1=3t

User Mschmidt
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2 Answers

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Answer:

No real solutions, t = (3 + i* sqrt(15))/12 t = (3-i sqrt(15))/12

Step-by-step explanation:

6t^2-3t+1= 0

delta will be equal to -15 which means no real solutions

t will be equal to (3+-i*sqrt(15))/12

User Girish Thimmegowda
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2 votes

Final answer:

To solve the quadratic 6t^2 + 1 = 3t, convert it to 6t^2 - 3t + 1 = 0 and use the quadratic formula. With a = 6, b = -3, and c = 1, we find the discriminant is negative, indicating no real solutions, only complex ones.

Step-by-step explanation:

To solve the quadratic equation 6t^2 + 1 = 3t using the quadratic formula, we first need to rearrange the equation into standard form, which is at^2 + bt + c = 0. In this case, we subtract 3t from both sides to get 6t^2 - 3t + 1 = 0. Now the equation is in the correct format, with a = 6, b = -3, and c = 1.

Next, we apply the quadratic formula:

t = (-b ± √(b^2 - 4ac)) / (2a)

Substituting our values we get:

t = (3 ± √((-3)^2 - 4(6)(1))) / (2*6)

t = (3 ± √(9 - 24)) / 12

t = (3 ± √(-15)) / 12

Since the discriminant (b^2 - 4ac) is negative, this equation has no real solutions. It only has complex solutions because the square root of a negative number is imaginary.

User Nfelger
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