Solve +2{-3[(-2+3-3)^2*(4-1+2)]-2+7-10}+1

$$+ 2 \{ - 3 [ ( - 2 + 3 - 3 ) ^ { 2 } \cdot ( 4 - 1 + 2 ) ] - 2 + 7 - 10 \} + 1$$

Evaluate

$$-129$$

Solution Steps

Add -2 and 3 to get 1.

$$2\left(-3\left(1-3\right)^{2}\left(4-1+2\right)-2+7-10\right)+1$$

Subtract 3 from 1 to get -2.

$$2\left(-3\left(-2\right)^{2}\left(4-1+2\right)-2+7-10\right)+1$$

Calculate -2 to the power of 2 and get 4.

$$2\left(-3\times 4\left(4-1+2\right)-2+7-10\right)+1$$

Multiply -3 and 4 to get -12.

$$2\left(-12\left(4-1+2\right)-2+7-10\right)+1$$

Subtract 1 from 4 to get 3.

$$2\left(-12\left(3+2\right)-2+7-10\right)+1$$

Add 3 and 2 to get 5.

$$2\left(-12\times 5-2+7-10\right)+1$$

Multiply -12 and 5 to get -60.

$$2\left(-60-2+7-10\right)+1$$

Subtract 2 from -60 to get -62.

$$2\left(-62+7-10\right)+1$$

Add -62 and 7 to get -55.

$$2\left(-55-10\right)+1$$

Subtract 10 from -55 to get -65.

$$2\left(-65\right)+1$$

Multiply 2 and -65 to get -130.

$$-130+1$$

Add -130 and 1 to get -129.

$$-129$$

Factor

$$-129$$