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
[tex] x^2 + 11x - 4 = 0 [/tex]
i.e. now we're looking for the roots of the polynomial. To find them, we can use the following formula:
[tex] x_{1,2} = \frac{-b\pm\sqrt{b^2-4ac}}{2} [/tex]
where [tex] x_{1,2} [/tex] is a compact way to indicate both solutions [tex] x_1 [/tex] and [tex] x_2 [/tex], while [tex] a,b,c [/tex] are the coefficients of the quadratic equation, i.e. we consider the polynomial [tex] ax^2+bx+c [/tex].
So, in your case, we have [tex] a=1,\ \ b=11,\ \ c=-4 [/tex]
Plug those values into the formula to get
[tex] x_{1,2} = \frac{-11\pm\sqrt{121+16}}{2} = \frac{-11\pm\sqrt{137}}{2} [/tex]
So, the two solutions are
[tex] x_1 = \frac{-11+\sqrt{137}}{2} [/tex]
[tex] x_2 = \frac{-11-\sqrt{137}}{2} [/tex]
