First sketch the graphs on each side of the equation and approximate the roots of the equation by approximating where the graphs cross

Approximate values for the roots are `x=0.2` and `1.2`

The formula for Newton's method is

`x_(n+1) = x_n - f(x_n)/(f'(x_n))`

Now, `f(x) = (x^2-x+1) +(-5e^(-x^2)sinx)`

`implies`

`f'(x) = (2x - 1 )+(10xe^(-x^2)sinx -5e^(-x^2)cosx)`

Working on the first root `x approx 0.2` gives the sequence

x0 = 0.2

x1 = 0.17677893

x2 = 0.17721437

x3 = 0.17721451

x4 = 0.17721451 `implies` root is 0.17721451 to 8dp

Working on the second root `x approx 1.2` gives the sequence

x0 = 1.2

x1 = 1.16247348

x2 = 1.16224076

x3 = 1.16224075

x4 = 1.16224075 `implies` root is 1.16224075 to 8dp

**The roots are 0.17721451, 1.16224075 to 8dp**

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