The possible values of \(x\) for \(2\cdot x^{2}+6\cdot x = 36\) are \(x_{1} = 3\) and \(x_{2} = -6\).
The expression \(2\cdot x^{2}+6\cdot x = 36\) is equivalent to the expression \(2\cdot x^{2}+6\cdot x -36 = 0\), which represents a second order polynomial, that is, a polynomial of the form:
\(y = a\cdot x^{2}+b\cdot x + c\) (1)
Where \(a,\,b,\,c\) are coefficients.
The possible values of x represents all possible roots of x, which are found analytically by the quadratic formula:
\(x = \frac{-b\pm \sqrt{b^{2}-4\cdot a\cdot c}}{2\cdot a}\) (2)
If we know that \(a = 2\), \(b = 6\) and \(c = -36\), then all possible values of \(x\) are, respectively:
\(x_{1} = 3\), \(x_{2} = -6\)
The possible values of \(x\) for \(2\cdot x^{2}+6\cdot x = 36\) are \(x_{1} = 3\) and \(x_{2} = -6\).
We kindly invite to check this question on second order polynomials: https://brainly.com/question/24356198
