Find all real solutions of the equation. (if there is no real solution, enter no real solution.) 35x2 12x − 36

Important:  35x2 12x − 36 is not an equation; to obtain an equation, you must set 35x2 12x − 36 = to 0:

35x^2 + 12x − 36 = 0.  Also, "the square of x" must be represented by "x^2," and you must insert either the + or the - sign in front of the term 12x.

I am assuming that you meant   35x^2 + 12x − 36 = 0.

If that is indeed the case, then a = 35, b = 12 and c = -36

and the discriminant is b^2 - 4ac, or (12)^2 - 4(35)(-36) = 5184.

Since the discriminant is positive, this equation has 2 real, unequal solutions.  They are
        -12 plus or minus sqrt(5184)
x =
                         2(35)

First subtract 10 from both side
y^2+6y-16=0
(y+8)(y-2)=0
-8 and 2