What is the real roots of this equation 1. x^2(x-9)(2x+1)=0 2. (x+4) (x^2-x+3)=0 3. (x+8) (x-7) (x^2-2x+5)=o ? and pls show me the process.

1.) x²(x-9)(2x +1) = 0
It means  x² = 0 or (x-9) = 0 or (2x +1) =0.    Solving each.
x = 0,    x = 0 +9, x = 9        2x +1 = 0   2x = 0 -1    2x = - 1,  x = -1/2
x = 0 or 9 or -1/2

2) (x + 4)(x² - x +3) = 0 
It means (x+4) = 0  or (x² - x +3) = 0
for:  x + 4 =0
         x = 0-4
          x = -4

for: x² - x +3 = 0. This is a quadratic equation. We would use quadratic formula:
Comparing to ax^2 + bx + c = 0, and using the quadratic formula:

x =  (-b + sqrt(b^2 - 4ac) / 2a
a = 1,  b = -1,  c = 3.
x =  (- -1  +  Square Root ( -1 ^ 2 - 4 * 1 * 3)) /  (2 *1))
x =  (1  +  Square Root (1 - 12) ) / 2  Imaginary roots.

x = -4 only real solution

3) (x+8)(x-7)(x²-2x+5) =0

x+8 = 0 or x-7 = 0 or  x²-2x+5 = 0
x = 0-8      x= 0 +7
x = -8        x = 7

x²-2x+5 = 0
x =  (-b + sqrt(b^2 - 4ac) / 2a
a = 1,  b = -2,  c = 5.
x =  (- -2  +  Square Root ( -2 ^ 2 - 4 * 1 * 5)) /  (2 *1)
x =  (1  +  Square Root (4 - 20) ) / 2  Imaginary roots.

x = -8, 7  the only real solutions