How to solve 2-3[1-5+3(-2)]

1. Factor the equation (x + 2)(x + 3) = 0. 2. Set each factor equal to zero: x + 2 = 0 and x + 3 = 0. 3. Solve for x: x = -2 and x = -3.

Solve 2-3[1-5+3(-2)] | Equatix

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3(r+2s)=2t-4

a⋅(b-2)=3b

3(r+2s)=2t-4

a⋅(b-2)=3b

Question

$$2-3[1-5+3(-2)]$$

Evaluate

$32$

Solution Steps

Subtract $5$ from $1$ to get $-4$.

$$2-3\left(-4+3\left(-2\right)\right)$$

Multiply $3$ and $-2$ to get $-6$.

$$2-3\left(-4-6\right)$$

Subtract $6$ from $-4$ to get $-10$.

$$2-3\left(-10\right)$$

Multiply $3$ and $-10$ to get $-30$.

$$2-\left(-30\right)$$

The opposite of $-30$ is $30$.

$$2+30$$

Add $2$ and $30$ to get $32$.

$$32$$

Show Solution

Factor

$2^{5}$

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