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y2-20y+99=0
We add all the numbers together, and all the variables
y^2-20y+99=0
a = 1; b = -20; c = +99;
Δ = b2-4ac
Δ = -202-4·1·99
Δ = 4
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{4}=2$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-2}{2*1}=\frac{18}{2} =9 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+2}{2*1}=\frac{22}{2} =11 $
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