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(20+y)y=220
We move all terms to the left:
(20+y)y-(220)=0
We add all the numbers together, and all the variables
(y+20)y-220=0
We multiply parentheses
y^2+20y-220=0
a = 1; b = 20; c = -220;
Δ = b2-4ac
Δ = 202-4·1·(-220)
Δ = 1280
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}$$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1280}=\sqrt{256*5}=\sqrt{256}*\sqrt{5}=16\sqrt{5}$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-16\sqrt{5}}{2*1}=\frac{-20-16\sqrt{5}}{2} $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+16\sqrt{5}}{2*1}=\frac{-20+16\sqrt{5}}{2} $
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