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=-16Y^2+32Y+12
We move all terms to the left:
-(-16Y^2+32Y+12)=0
We get rid of parentheses
16Y^2-32Y-12=0
a = 16; b = -32; c = -12;
Δ = b2-4ac
Δ = -322-4·16·(-12)
Δ = 1792
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}$$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1792}=\sqrt{256*7}=\sqrt{256}*\sqrt{7}=16\sqrt{7}$$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-32)-16\sqrt{7}}{2*16}=\frac{32-16\sqrt{7}}{32} $$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-32)+16\sqrt{7}}{2*16}=\frac{32+16\sqrt{7}}{32} $
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