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2y^2-5y-13=0
a = 2; b = -5; c = -13;
Δ = b2-4ac
Δ = -52-4·2·(-13)
Δ = 129
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{-(-5)-\sqrt{129}}{2*2}=\frac{5-\sqrt{129}}{4} $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+\sqrt{129}}{2*2}=\frac{5+\sqrt{129}}{4} $
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