What is expand 2(x+Δx)^2-3(x+Δx) ?

The solution to expand 2(x+Δx)^2-3(x+Δx) is 2x^2+4Δx^2+2Δ^2x^2-3x-3Δx

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expand 2(x+Δx)^2-3(x+Δx)

expand

2 (x + Δ x)^{2} - 3 (x + Δ x)

2 x^{2} + 4Δ x^{2} + 2 Δ^{2} x^{2} - 3 x - 3Δ x

Solution steps

2 (x + Δ x)^{2} - 3 (x + Δ x)

Expand

- 3 (x + Δ x) : - 3 x - 3Δ x

= 2 (x + Δ x)^{2} - 3 x - 3Δ x

Simplify

2 (x + Δ x)^{2} - 3 x - 3Δ x : 2 x^{2} + 4Δ x^{2} + 2 Δ^{2} x^{2} - 3 x - 3Δ x

= 2 x^{2} + 4Δ x^{2} + 2 Δ^{2} x^{2} - 3 x - 3Δ x

\frac{d}{d t} ((x) (4 x + 7 y))

(y cos (x)) (x + d x)

\frac{1}{2} (1 - e^{- 10 t})

\frac{n \cdot ( n - 3 )}{2}

\frac{( x - 3 )}{( x - 4 ) ( x - 2 )}

What is expand 2(x+Δx)^2-3(x+Δx) ?
The solution to expand 2(x+Δx)^2-3(x+Δx) is 2x^2+4Δx^2+2Δ^2x^2-3x-3Δx