64a^2+8=17

Solution for 64a^2+8=17 equation:

64a^2+8=17

We move all terms to the left:

64a^2+8-(17)=0

We add all the numbers together, and all the variables

64a^2-9=0

a = 64; b = 0; c = -9;
Δ = b2-4ac
Δ = 02-4·64·(-9)
Δ = 2304
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2304}=48$
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-48}{2*64}=\frac{-48}{128}
=-3/8
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+48}{2*64}=\frac{48}{128}
=3/8
$