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10-(2y-6)y=8
We move all terms to the left:
10-(2y-6)y-(8)=0
We add all the numbers together, and all the variables
-(2y-6)y+2=0
We multiply parentheses
-2y^2+6y+2=0
a = -2; b = 6; c = +2;
Δ = b2-4ac
Δ = 62-4·(-2)·2
Δ = 52
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}$$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{52}=\sqrt{4*13}=\sqrt{4}*\sqrt{13}=2\sqrt{13}$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-2\sqrt{13}}{2*-2}=\frac{-6-2\sqrt{13}}{-4} $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+2\sqrt{13}}{2*-2}=\frac{-6+2\sqrt{13}}{-4} $
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