Solve the quadratic (x+57)/6=x^2

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