Solve x^2+4x-5=0

Two approaches.1.Note that the equation is a quadratic equation, and can thus be solved per the quadratic formula, in which the solutions are given to be (-b+-sqrt(b^2-4ac))/2a.Plugging in the values from the quadrartic, one obtains (-4+-sqrt(36))/2, i.e. x equals 1 and -5.2.Solve by collecting terms and expressing in the form (ax+b)(cx+d)=0. One notices that the following must hold.cd=-5ad+cb=4ac=1While the former approach is generally easier, as many students will try the latter with trial and error, the latter yields nice numbers in this instance: (x+5)(x-1)=0. The solutions to the quadratic occur when the RHS=LHS, i.e. when one of the brackets evaluates to 0. Therefore, one obtains x equals -5 and 1.