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=2Y^2-16Y
We move all terms to the left:
-(2Y^2-16Y)=0
We get rid of parentheses
-2Y^2+16Y=0
a = -2; b = 16; c = 0;
Δ = b2-4ac
Δ = 162-4·(-2)·0
Δ = 256
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}$$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{256}=16$$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-16}{2*-2}=\frac{-32}{-4} =+8 $$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+16}{2*-2}=\frac{0}{-4} =0 $
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