Trinomials X^2+2x-4=0?

Question Number 1 :
For this equation x^2 + 2*x - 4 = 0 , answer the following questions :
A. Find the roots using Quadratic Formula !

Answer Number 1 :
The equation x^2 + 2*x - 4 = 0 is already in a*x^2+b*x+c=0 form.
As the value is already arranged in a*x^2+b*x+c=0 form, we get the value of a = 1, b = 2, c = -4.

1A. Find the roots using Quadratic Formula !
Remember the formula,
x1 = (-b+sqrt(b^2-4*a*c))/(2*a) and x2 = (-b-sqrt(b^2-4*a*c))/(2*a)
We had know that a = 1, b = 2 and c = -4,
then the value a,b and c in the abc formula, can be subtituted.
So we get x1 = (-(2) + sqrt( (2)^2 - 4 * (1)*(-4)))/(2*1) and x2 = (-(2) - sqrt( (2)^2 - 4 * (1)*(-4)))/(2*1)
Which is the same with x1 = ( -2 + sqrt( 4+16))/(2) and x2 = ( -2 - sqrt( 4+16))/(2)
Which is the same with x1 = ( -2 + sqrt( 20))/(2) and x2 = ( -2 - sqrt( 20))/(2)
We can get x1 = ( -2 + 4.47213595499958 )/(2) and x2 = ( -2 - 4.47213595499958 )/(2)
So we got the answers as x1 = 1.23606797749979 and x2 = -3.23606797749979

x^2 + 2x - 4 = 0
x = [-b ±√(b^2 - 4ac)]/(2a)

a = 1
b = 2
c = -4

x = [-2 ±√(4 + 16)]/2
x = [-2 ±√20]/2
x = [-2 ±√(2^2 * 5)]/2
x = [-2 ±2√5]/2
x = -1 ±√5

∴ x = -1 ±√5

x = [-b ± sqrt(b^2-4ac) ] / 2a
-3.2 1.2

(x+1+\/''5'' )(x+1-\/''5'' )=0

x= -3.236 or 1.236


which are just:

- (sq rt 5) - 1
(sq rt 5) - 1

use the formula,

x = [-b ± sqrt(b^2-4ac) ] / 2a

use the formula,

x = [-b ± sqrt(b^2-4ac) ] / 2a

in your case a=1, b=2 , c=-4

pluck it in and you will get the answer

x = -1+sqrt5 or -1 - sqrt5