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(50-y)y=50
We move all terms to the left:
(50-y)y-(50)=0
We add all the numbers together, and all the variables
(-1y+50)y-50=0
We multiply parentheses
-1y^2+50y-50=0
a = -1; b = 50; c = -50;
Δ = b2-4ac
Δ = 502-4·(-1)·(-50)
Δ = 2300
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}$$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{2300}=\sqrt{100*23}=\sqrt{100}*\sqrt{23}=10\sqrt{23}$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(50)-10\sqrt{23}}{2*-1}=\frac{-50-10\sqrt{23}}{-2} $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(50)+10\sqrt{23}}{2*-1}=\frac{-50+10\sqrt{23}}{-2} $
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