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(8y-4)8y=6
We move all terms to the left:
(8y-4)8y-(6)=0
We multiply parentheses
64y^2-32y-6=0
a = 64; b = -32; c = -6;
Δ = b2-4ac
Δ = -322-4·64·(-6)
Δ = 2560
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}$$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{2560}=\sqrt{256*10}=\sqrt{256}*\sqrt{10}=16\sqrt{10}$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-32)-16\sqrt{10}}{2*64}=\frac{32-16\sqrt{10}}{128} $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-32)+16\sqrt{10}}{2*64}=\frac{32+16\sqrt{10}}{128} $
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