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=40+20Y-5Y^2
We move all terms to the left:
-(40+20Y-5Y^2)=0
We get rid of parentheses
5Y^2-20Y-40=0
a = 5; b = -20; c = -40;
Δ = b2-4ac
Δ = -202-4·5·(-40)
Δ = 1200
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}$$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1200}=\sqrt{400*3}=\sqrt{400}*\sqrt{3}=20\sqrt{3}$$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-20\sqrt{3}}{2*5}=\frac{20-20\sqrt{3}}{10} $$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+20\sqrt{3}}{2*5}=\frac{20+20\sqrt{3}}{10} $
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