Solve the quadratic (x+57)/6=x^2
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We have to solve (x+57)/6=x^2
(x+57)/6=x^2
=> x + 57 = 6x^2
=> 6x^2 - x - 57 = 0
The roots of a quadratic equation ax^2 + bx + c = 0 are given by [-b + sqrt (b^2 - 4ac)]/2a and [-b + sqrt (b^2 - 4ac)]/2a.
Here a = 6 , b = -1 and c = -57.
Substituting we get
x1 = [1 + sqrt(1 + 24*57)]/12
=> x1 = 1/12 + sqrt 1369 / 12
=> x1 = 1/12 + 37/12
=> x1 = 38/12
=> x1 = 19/6
x2 = 1/12 - sqrt 1369 / 12
=> x2 = 1/12 - 37/12
=> x2 = -36/12
=> x2 = -3
Therefore x is equal to -3 and 19/6
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First, we'll multiply by 6 both sides:
x + 57 = 6x^2
Now, we'll subtract x + 57 both sides and we'll apply symmetric property:
6x^2 - x - 57 = 0
We'll apply the quadratic formula:
x1 = [1 + sqrt(1 + 24*57)]/12
x1 = (1 + sqrt1369)/12
x1 = (1 + 37)/12
x1 = 38/12
x1 = 19/6
x2 = (1-37)/12
x2 = -36/12
x2 = -3
The solutions of the equation are: {-3 ; 19/6}.
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