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y2-3y-5=0
We add all the numbers together, and all the variables
y^2-3y-5=0
a = 1; b = -3; c = -5;
Δ = b2-4ac
Δ = -32-4·1·(-5)
Δ = 29
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{-(-3)-\sqrt{29}}{2*1}=\frac{3-\sqrt{29}}{2} $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{29}}{2*1}=\frac{3+\sqrt{29}}{2} $
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