64y2+7=32

Solution for 64y2+7=32 equation:

64y^2+7=32

We move all terms to the left:

64y^2+7-(32)=0

We add all the numbers together, and all the variables

64y^2-25=0

a = 64; b = 0; c = -25;
Δ = b2-4ac
Δ = 02-4·64·(-25)
Δ = 6400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{6400}=80$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-80}{2*64}=\frac{-80}{128}
=-5/8
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+80}{2*64}=\frac{80}{128}
=5/8
$