Solve (a+2)^2-5(3-2a)+(2a-3)^2

$$( a + 2 ) ^ { 2 } - 5 ( 3 - 2 a ) + ( 2 a - 3 ) ^ { 2 }$$

Evaluate

$$5a^{2}+2a-2$$

Solution Steps

Use binomial theorem \left(p+q\right)^{2}=p^{2}+2pq+q^{2} to expand \left(a+2\right)^{2}.

$$a^{2}+4a+4-5\left(3-2a\right)+\left(2a-3\right)^{2}$$

Use the distributive property to multiply -5 by 3-2a.

$$a^{2}+4a+4-15+10a+\left(2a-3\right)^{2}$$

Subtract 15 from 4 to get -11.

$$a^{2}+4a-11+10a+\left(2a-3\right)^{2}$$

Combine 4a and 10a to get 14a.

$$a^{2}+14a-11+\left(2a-3\right)^{2}$$

Use binomial theorem \left(p-q\right)^{2}=p^{2}-2pq+q^{2} to expand \left(2a-3\right)^{2}.

$$a^{2}+14a-11+4a^{2}-12a+9$$

Combine a^{2} and 4a^{2} to get 5a^{2}.

$$5a^{2}+14a-11-12a+9$$

Combine 14a and -12a to get 2a.

$$5a^{2}+2a-11+9$$

Add -11 and 9 to get -2.

$$5a^{2}+2a-2$$

Expand

$$5a^{2}+2a-2$$

Solution Steps

Use binomial theorem \left(p+q\right)^{2}=p^{2}+2pq+q^{2} to expand \left(a+2\right)^{2}.

$$a^{2}+4a+4-5\left(3-2a\right)+\left(2a-3\right)^{2}$$

Use the distributive property to multiply -5 by 3-2a.

$$a^{2}+4a+4-15+10a+\left(2a-3\right)^{2}$$

Subtract 15 from 4 to get -11.

$$a^{2}+4a-11+10a+\left(2a-3\right)^{2}$$

Combine 4a and 10a to get 14a.

$$a^{2}+14a-11+\left(2a-3\right)^{2}$$

Use binomial theorem \left(p-q\right)^{2}=p^{2}-2pq+q^{2} to expand \left(2a-3\right)^{2}.

$$a^{2}+14a-11+4a^{2}-12a+9$$

Combine a^{2} and 4a^{2} to get 5a^{2}.

$$5a^{2}+14a-11-12a+9$$

Combine 14a and -12a to get 2a.

$$5a^{2}+2a-11+9$$

Add -11 and 9 to get -2.

$$5a^{2}+2a-2$$