Question

Can someone help me solve these quadratics by completing the square??
x^2+2x-7=0 and -x^2+6x+10=0

Answer 1

@TuringTest ? @tcarroll010 ?

Answer 2

@Algebraic! ? @saifoo.khan ?

Answer 3

As an example I'll do the first..\[x^2+2x-7=0\]\[x^2+2x+1-1-7=0\qquad\text{You divide the coefficient of x and square the new}\]\[\text{ term
to get the new value 1}\]\[\qquad\]\[x^2+2x+1-8=0\]\[(x+1)^2-8=0\]\[(x+1)^2=8\]\[x+1=\pm\sqrt{8}\]\[x=-1\pm\sqrt{8}\]\[\text{So you get two values..}\]\[x=-1+\sqrt{8}\qquad\text{or}\qquad x=-1-\sqrt{8}\]