Question 208231
r^2+3r=5
r^2+3r-5=0
Using the quadratic equation {{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}} we get:
r=(-3+-sqrt[3^2-4*1*-5])/281
r=(-3+-sqrt[9+20])/2
r=(-3+-sqrt29)/2
r=(-3+-5.385)/2
r=(-3+5.385)/2
r=1.1925 ans.
r=-3-5.385/2
r=-8.385/2
r=-4.19 ans.