This is a quadratic equation, i.e. an equation involving a polynomial of degree 2. To solve them, you must rearrange them first, so that all terms are on the same side, so we get

i.e. now we're looking for the roots of the polynomial. To find them, we can use the following formula:

where
is a compact way to indicate both solutions
and
, while
are the coefficients of the quadratic equation, i.e. we consider the polynomial
.
So, in your case, we have 
Plug those values into the formula to get

So, the two solutions are


Answer:
No Solutions
Step-by-step explanation:
In a triangle, the sum of the angles has to be 180 degrees. It also is impossible to find the length of sides without at least one side, since the range of lengths is practically infinite. There are no solutions to this problem.
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The answer to this problem is 9 hopefully this helps you!