Solve -(3n^2-2n+4)-[2n^2-(n^2+n+3)]

$$- ( 3 n ^ { 2 } - 2 n + 4 ) - [ 2 n ^ { 2 } - ( n ^ { 2 } + n + 3 ) ]$$

Evaluate

$$-4n^{2}+3n-1$$

Solution Steps

To find the opposite of 3n^{2}-2n+4, find the opposite of each term.

$$-3n^{2}+2n-4-\left(2n^{2}-\left(n^{2}+n+3\right)\right)$$

To find the opposite of n^{2}+n+3, find the opposite of each term.

$$-3n^{2}+2n-4-\left(2n^{2}-n^{2}-n-3\right)$$

Combine 2n^{2} and -n^{2} to get n^{2}.

$$-3n^{2}+2n-4-\left(n^{2}-n-3\right)$$

To find the opposite of n^{2}-n-3, find the opposite of each term.

$$-3n^{2}+2n-4-n^{2}+n+3$$

Combine -3n^{2} and -n^{2} to get -4n^{2}.

$$-4n^{2}+2n-4+n+3$$

Combine 2n and n to get 3n.

$$-4n^{2}+3n-4+3$$

Add -4 and 3 to get -1.

$$-4n^{2}+3n-1$$

Expand

$$-4n^{2}+3n-1$$

Solution Steps

To find the opposite of 3n^{2}-2n+4, find the opposite of each term.

$$-3n^{2}+2n-4-\left(2n^{2}-\left(n^{2}+n+3\right)\right)$$

To find the opposite of n^{2}+n+3, find the opposite of each term.

$$-3n^{2}+2n-4-\left(2n^{2}-n^{2}-n-3\right)$$

Combine 2n^{2} and -n^{2} to get n^{2}.

$$-3n^{2}+2n-4-\left(n^{2}-n-3\right)$$

To find the opposite of n^{2}-n-3, find the opposite of each term.

$$-3n^{2}+2n-4-n^{2}+n+3$$

Combine -3n^{2} and -n^{2} to get -4n^{2}.

$$-4n^{2}+2n-4+n+3$$

Combine 2n and n to get 3n.

$$-4n^{2}+3n-4+3$$

Add -4 and 3 to get -1.

$$-4n^{2}+3n-1$$