+118613 Show that x^2+90x-1200=0
I can't cause it doesn't. I mean it might sometimes but it very rarely would.
I think the question is supposed to be
Find when x^2+90x-1200=0
Or
Solve for x: x^2+90x-1200=0
I'd use the quadratic formula to solve this.
\(\triangle =b^2-4ac \\ \triangle = 8100+4*1200\\ \triangle =8100+4800\\ \triangle =12900\\ x=\frac{-90\pm\sqrt{12900}}{2}\\ x=\frac{-90\pm10\sqrt{129}}{2}\\ x=-45\pm5\sqrt{129}\\\)
