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=-2(Y+1)(Y+5)
We move all terms to the left:
-(-2(Y+1)(Y+5))=0
We multiply parentheses ..
-(-2(+Y^2+5Y+Y+5))=0
We calculate terms in parentheses: -(-2(+Y^2+5Y+Y+5)), so:We get rid of parentheses
-2(+Y^2+5Y+Y+5)
We multiply parentheses
-2Y^2-10Y-2Y-10
We add all the numbers together, and all the variables
-2Y^2-12Y-10
Back to the equation:
-(-2Y^2-12Y-10)
2Y^2+12Y+10=0
a = 2; b = 12; c = +10;
Δ = b2-4ac
Δ = 122-4·2·10
Δ = 64
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{64}=8$$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-8}{2*2}=\frac{-20}{4} =-5 $$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+8}{2*2}=\frac{-4}{4} =-1 $
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