Solve 2(1-t)^2-3(1-t)+10

$$2 ( 1 - t ) ^ { 2 } - 3 ( 1 - t ) + 10$$

Evaluate

$$2t^{2}-t+9$$

Solution Steps

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(1-t\right)^{2}.

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

Use the distributive property to multiply 2 by 1-2t+t^{2}.

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

Use the distributive property to multiply -3 by 1-t.

$$2-4t+2t^{2}-3+3t+10$$

Subtract 3 from 2 to get -1.

$$-1-4t+2t^{2}+3t+10$$

Combine -4t and 3t to get -t.

$$-1-t+2t^{2}+10$$

Add -1 and 10 to get 9.

$$9-t+2t^{2}$$

Expand

$$2t^{2}-t+9$$

Solution Steps

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(1-t\right)^{2}.

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

Use the distributive property to multiply 2 by 1-2t+t^{2}.

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

Use the distributive property to multiply -3 by 1-t.

$$2-4t+2t^{2}-3+3t+10$$

Subtract 3 from 2 to get -1.

$$-1-4t+2t^{2}+3t+10$$

Combine -4t and 3t to get -t.

$$-1-t+2t^{2}+10$$

Add -1 and 10 to get 9.

$$9-t+2t^{2}$$