Newton–Raphson method is used to compute a root of the equation x2 + 5x − 10 =…

We apply the Newton–Raphson iteration formula:

xₙ₊₁ = xₙ − f(xₙ) / f′(xₙ)

Here f(x) = x² + 5x − 10, so the derivative is f′(x) = 2x + 5.

Step 1 — Evaluate f and f′ at the initial value x₀ = 1:

f(1) = (1)² + 5(1) − 10 = 1 + 5 − 10 = −4

f′(1) = 2(1) + 5 = 7

Step 2 — Apply the formula for the first iteration:

x₁ = 1 − [ f(1) / f′(1) ] = 1 − (−4 / 7) = 1 + 4/7

x₁ = 1 + 0.5714… ≈ 1.571

Therefore, after one iteration the approximate root is 1.571 (rounded to three decimal places).