How do you solve #x^2 + 2x-7=0# using the quadratic formula?

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Answer 1 · 2016-02-14T09:41:52

#color(green)(x=-1+-2sqrt2#

#color(blue)(x^2+2x-7#

This equation is a Quadratic equation (in form #ax^2+bx+c=0#)

#color(brown)(x=(-b+-sqrt(b^2-4ac))/(2a)#

In this case #color(red)(a=1,b=2,c=-7#

Substitute the values:

#rarrx=color(orange)((-2+-sqrt(2^2-4(1)(-7)))/(2(1))#

#rarrx=color(orange)((-2+-sqrt(4-(-28)))/(2)#

#rarrx=color(orange)((-2+-sqrt(4+28))/2#

#rarrx=color(orange)((-2+-sqrt32)/2#

#rarrx=color(orange)((-2+-sqrt(16*2))/2#

#rarrx=color(orange)((-2+-4sqrt2)/2#

#rarrx=color(orange)(-2/2+-(4sqrt2)/2#

#rarrx=color(orange)(-cancel(2)/cancel(2)+-(cancel4sqrt2)/cancel2#

#rArrcolor(green)(x=-1+-2sqrt2#