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=3(Y-5)(Y-2)
We move all terms to the left:
-(3(Y-5)(Y-2))=0
We multiply parentheses ..
-(3(+Y^2-2Y-5Y+10))=0
We calculate terms in parentheses: -(3(+Y^2-2Y-5Y+10)), so:We get rid of parentheses
3(+Y^2-2Y-5Y+10)
We multiply parentheses
3Y^2-6Y-15Y+30
We add all the numbers together, and all the variables
3Y^2-21Y+30
Back to the equation:
-(3Y^2-21Y+30)
-3Y^2+21Y-30=0
a = -3; b = 21; c = -30;
Δ = b2-4ac
Δ = 212-4·(-3)·(-30)
Δ = 81
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{81}=9$$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(21)-9}{2*-3}=\frac{-30}{-6} =+5 $$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(21)+9}{2*-3}=\frac{-12}{-6} =+2 $
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