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y2+2.503y-1.733=0
We add all the numbers together, and all the variables
y^2+2.503y-1.733=0
a = 1; b = 2.503; c = -1.733;
Δ = b2-4ac
Δ = 2.5032-4·1·(-1.733)
Δ = 13.197009
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}$$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2.503)-\sqrt{13.197009}}{2*1}=\frac{-2.503-\sqrt{13.197009}}{2} $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2.503)+\sqrt{13.197009}}{2*1}=\frac{-2.503+\sqrt{13.197009}}{2} $
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