Can someone please factorise 5x²- 22x + 80=0?

5x^2 - 22x + 80 = 0
x = [-b ± √(b^2 - 4ac)]/(2a)

a = 5
b = -22
c = 80

x = [22 ± √(484 - 1600)]/10
x = [22 ± √(-1116)]/10
x = [22 ± √(i^2 * 6^2 * 31)]/10
x = [22 ± 6i√31]/10
x = [11 ± 3i√31]/5

∴ x = [11 ± 3i√31]/5

No factors. Roots are complex conjugates

This will not factor without imaginary values so we need to use quadratic formula instead.

a = 5
b = -22
c = 80

x = [ -(-22) ± √(-22^ - 4 * 5 * 80)] / (2 * 5)

x = [22 ± √(484 - 1600) ] /10

x = (22 ± √-1116) /10

x = 2.2 ± 6i√31

You cannot factorise as the discriminant is not a square number.

b^2-4ac = square to factorise.

5x²- 22x + 80=0 >>>> a= 5 , b= -22, c=80

=> (-22^2) -4x5x80

=> 484 - 4 x 5 x 80

= -1116

Since -1116 is not square it does not factorise.

use the formula for finding complex roots

x = {-b ± √( b² - 4ac)} / 2a

Not factorable.

Not factorisable without imaginaries