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y2+53y+630=0
We add all the numbers together, and all the variables
y^2+53y+630=0
a = 1; b = 53; c = +630;
Δ = b2-4ac
Δ = 532-4·1·630
Δ = 289
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{289}=17$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(53)-17}{2*1}=\frac{-70}{2} =-35 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(53)+17}{2*1}=\frac{-36}{2} =-18 $
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