What is expand \exists x(x+2=1) ?

The solution to expand \exists x(x+2=1) is \exists x^2+2\exists x=1

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expand \exists x(x+2=1)

expand

\exists x (x + 2 = 1)

\exists x^{2} + 2\exists x = 1

Solution steps

\exists x (x + 2 = 1)

= \exists x (x + 2 \cdot 1 =)

Apply the distributive law:

a (b + c) = ab + a c

a = \exists x, b = x, c = 2 = 1

\exists xx + \exists x \cdot 2 = 1

\exists xx + 2\exists x = 1

Expand

\exists xx + 2\exists x : \exists x^{2} + 2\exists x

\exists x^{2} + 2\exists x = 1

4 (3 x + 2)^{2} + 4

\frac{( 7 y - 6 ) ^{2}}{2}

(a + a^{^{'}}) b

\frac{2}{9} (3 e^{3 x} x - e^{3 x}) + C

\frac{2 pq}{( 1 - q ) ^{3}}

What is expand \exists x(x+2=1) ?
The solution to expand \exists x(x+2=1) is \exists x^2+2\exists x=1