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2a^2-2a=16
We move all terms to the left:
2a^2-2a-(16)=0
a = 2; b = -2; c = -16;
Δ = b2-4ac
Δ = -22-4·2·(-16)
Δ = 132
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}$$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{132}=\sqrt{4*33}=\sqrt{4}*\sqrt{33}=2\sqrt{33}$$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-2\sqrt{33}}{2*2}=\frac{2-2\sqrt{33}}{4} $$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2\sqrt{33}}{2*2}=\frac{2+2\sqrt{33}}{4} $
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