Using the quadratic formula to solve 5x = 6x2 – 3, what are the values of x? StartFraction 5 plus-or-minus 3 StartRoot 11 EndRoot Over 12 EndFraction StartFraction 5 plus-or-minus StartRoot 97 EndRoot Over 12 EndFraction StartFraction 5 plus-or-minus StartRoot 47 EndRoot Over 12 EndFraction StartFraction negative 5 plus-or-minus StartRoot 97 EndRoot Over 12 EndFraction

Using the quadratic formula to solve 5x = 6x^2 – 3, what are the values of x?

5x = 6x^2 – 3

Subtract 5x from both sides:

0 = 6x^2 – 5x – 3

a = 6, b = -5, c = -3

x = (-b ± √(b^2 - 4ac))/(2a)

x = (-(-5) ± √((-5)^2 - 4(6)(-3)))/(2(6))

x = (5 ± √(25 + 72))/12

x = (5 ± √(97))/12

Step-by-step explanation:

The answer is B

Step-by-step explanation:

1/800 is the right answer

1. 3. can also be expressed as (22/7)

2. -4.5

3. 3.6

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