(4y-5)(4y+5)=0

Solution for (4y-5)(4y+5)=0 equation:

(4y-5)(4y+5)=0

We use the square of the difference formula

16y^2-25=0

a = 16; b = 0; c = -25;
Δ = b2-4ac
Δ = 02-4·16·(-25)
Δ = 1600
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1600}=40$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-40}{2*16}=\frac{-40}{32}
=-1+1/4
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+40}{2*16}=\frac{40}{32}
=1+1/4
$