How do you solve (2y-3)(2y+3) = 16?

Feb 9, 2016

Solution is y=5/2 or −5/2

Explanation:

$\left(2 y - 3\right) \cdot \left(2 y + 3\right) = 16$

is $4 {y}^{2} - 9 = 16$

i.e. $4 {y}^{2} = 16 + 9 = 25$

i.e. $4 {y}^{2} - 25 = 0$

Factorizing this it becomes

$\left(2 y - 5\right) \cdot \left(2 y + 5\right) = 0$

Hence either $\left(2 y - 5\right) = 0$ which gives $y = \frac{5}{2}$

or

$\left(2 y + 5\right) = 0$ which gives $y = - \frac{5}{2}$

Hence solution is $y = \frac{5}{2} \mathmr{and} - \frac{5}{2}$