How do you solve #x^2+2x-5=0#?

Use the improved quadratic formula (Google, Yahoo Search)
#D = d^2 = b^2 - 4ac = 4 + 20 = 24# --> #d = +- 2sqrt6#
There are 2 real roots:
#x = - 2/2 +- (2sqrt6)/2 = -1 +- sqrt6#

#color(blue)(x^2+2x-5=0#

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

Use Quadratic equation

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

Where

#color(red)(a=1,b=2,c=-5#

#rarrx=(-2+-sqrt(2^2-4(1)(-5)))/(2(1))#

#rarrx=(-2+-sqrt(4-4(-5)))/(2)#

#rarrx=(-2+-sqrt(4-(-20)))/(2)#

#rarrx=(-2+-sqrt(4-(-20)))/(2)#

#rarrx=(-2+-sqrt(4+20))/(2)#

#rarrx=(-2+-sqrt(24))/(2)#

#rarrx=(-2+-sqrt(4*6))/(2)#

#rarrx=(-2+-2sqrt(6))/(2)#

#rarrx=(-cancel2+-cancel2sqrt(6))/(cancel2)#

#color(green)(rArrx=-1+-sqrt6#

Another approach would be to complete the square. It is another form of the standard #x=(-b+-sqrt(b^2-4ac))/(2a)# .

#color(brown)("It has to be another form otherwise you would not be")##color(brown)("able to solve for x when y=0.")#

Given:#" "x^2+2x-5=0#

Note that standard form of #" "y=ax^2+bx+c# becomes

#y=a(x+b/(2a))^2+c+k" "# where #k# is a corrective constant

#=>y= (x+1)^2-5+k#

The error comes from #(b/2)^2 -> (+2/2)^2#

So #(+2/2)^2+k=0" "=> k=-1# giving

#y=(x+1)^2-6#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#y+6=(x+1)^2#

Taking the square root of both sides

#sqrt(y+6)=x+1#

But for this question #y=0# giving

#+-sqrt(6)-1=x#

#color(blue)(=> x~~ 1.449" or "x~~-3.449" to 3 decimal places")#