How do you solve x²-x-4=0 ?

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

a = 1
b = -1
c = -4

x = [1 ±√(1 + 16)]/2
x = [1 ±√17]/2
x = [1 ±4.12]/2 (approx.)

x = [1 + 4.12]/2
x = 5.12/2
x = 2.56

x = [1 - 4.12]/2
x = -3.12/2
x = -1.56

∴ x = -1.56 , 2.56

as total if we have this : a*x^2+b*x+c=0 the we have :
X= (-b (-or+)sqrt(b^2-4*a*c))/(2*a)
so in this case we have:X=(1 +or- sqrt(17))/2

in a quadratic equation ,ax²+bx+c=0, x=(-b+/-under root(b²-4ac))/2a

x = [ - b ± √ (b ² - 4 a c ) ] / 2 a
x = [ 1 ± √ (1 + 16 ) ] / 2
x = [ 1 ± √17 ] / 2

x = [ - b ± √ (b ² - 4 a c ) ] / 2 a
x = [ 1 ± √ (1 + 16 ) ] / 2
x = [ 1 ± √17 ] / 2