What is the sum of the two solutions to the equation $54-15x-x^2=0$?
I want to find the values of x that make the equation true, so I need to solve for x. I could try to factor the equation, but it looks a bit messy, so I'll use the quadratic formula instead. The quadratic formula says that if I have an equation of the form $ax^2+bx+c=0$, then the solutions are $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$. In this case, $a=-1$, $b=-15$, and $c=54$, so I plug those values into the formula and get $x=\frac{-(-15)\pm\sqrt{(-15)^2-4(-1)(54)}}{2(-1)}$. I simplify the expression and get $x=\frac{15\pm\sqrt{225+216}}{-2}$. I compute the square root and get $x=\frac{15\pm\sqrt{441}}{-2}$. I notice that 441 is a perfect square, so I can simplify further and get $x=\frac{15\pm 21}{-2}$. I split the expression into two possible solutions and get $x=\frac{15+21}{-2}$ or $x=\frac{15-21}{-2}$. I simplify the fractions and get $x=-18$ or $x=3$. These are the two solutions to the equation, and the question asks for their sum. I add them together and get $(-18)+3=-15$. 