# algebra1

justaguide | Certified Educator

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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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giorgiana1976 | Student

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

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}.

check Approved by eNotes Editorial