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=(6Y+4)Y=5
We move all terms to the left:
-((6Y+4)Y)=0
We calculate terms in parentheses: -((6Y+4)Y), so:We get rid of parentheses
(6Y+4)Y
We multiply parentheses
6Y^2+4Y
Back to the equation:
-(6Y^2+4Y)
-6Y^2-4Y=0
a = -6; b = -4; c = 0;
Δ = b2-4ac
Δ = -42-4·(-6)·0
Δ = 16
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{16}=4$$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-4}{2*-6}=\frac{0}{-12} =0 $$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+4}{2*-6}=\frac{8}{-12} =-2/3 $
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