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16x^2=-8x+1
We move all terms to the left:
16x^2-(-8x+1)=0
We get rid of parentheses
16x^2+8x-1=0
a = 16; b = 8; c = -1;
Δ = b2-4ac
Δ = 82-4·16·(-1)
Δ = 128
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{128}=\sqrt{64*2}=\sqrt{64}*\sqrt{2}=8\sqrt{2}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-8\sqrt{2}}{2*16}=\frac{-8-8\sqrt{2}}{32} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+8\sqrt{2}}{2*16}=\frac{-8+8\sqrt{2}}{32} $
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