Question

Dada la ecuación (x+2)2+5x(x-1)=55
la solución positiva es

Answer 1

$(x+2)^2+5x(x-1)=55 \\ \\ (x+2)*(x+2)+5x(x-1)=55 \\ \\ x^2+2x+2x+4+5x^2-5x=55 \\ \\ x^2+5x^2+2x+2x-5x+4=55 \\ \\ 6x^2-x=55-4 \\ \\ 6x^2-x=51 \\ \\ 6x^2-x-51=0 \\ \\ \\ x= \frac{-b\pm \sqrt{b^2-4ac} }{2a} = \frac{1\pm \sqrt{(-1)^2-4*6*(-51)} }{2*6} = \frac{1\pm \sqrt{1+1224} }{12} = \frac{1\pm \sqrt{(1225} }{12} = \\ \\ = \frac{1\pm 35 }{12} = \left \{ {{x= \frac{36}{12};x=3 } \atop {x= \frac{-34}{12};x= -\frac{17}{6} }} \right.$

Como descartamos el valor negativo de x, entonces x=3.