Solve F=(4t-1)(4t+1)-(3t+2)^2

$$F = ( 4 t - 1 ) ( 4 t + 1 ) - ( 3 t + 2 ) ^ { 2 }$$

Solve for F

$$F=7t^{2}-12t-5$$

Solution Steps

Consider \left(4t-1\right)\left(4t+1\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 1.

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

Expand \left(4t\right)^{2}.

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

Calculate 4 to the power of 2 and get 16.

$$F=16t^{2}-1-\left(3t+2\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(3t+2\right)^{2}.

$$F=16t^{2}-1-\left(9t^{2}+12t+4\right)$$

To find the opposite of 9t^{2}+12t+4, find the opposite of each term.

$$F=16t^{2}-1-9t^{2}-12t-4$$

Combine 16t^{2} and -9t^{2} to get 7t^{2}.

$$F=7t^{2}-1-12t-4$$

Subtract 4 from -1 to get -5.

$$F=7t^{2}-5-12t$$

Solve for t

$$t=\frac{-\sqrt{7F+71}+6}{7}$$

$$t=\frac{\sqrt{7F+71}+6}{7}\text{, }F\geq -\frac{71}{7}$$