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2y^2-3y-65=0
a = 2; b = -3; c = -65;
Δ = b2-4ac
Δ = -32-4·2·(-65)
Δ = 529
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{529}=23$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-23}{2*2}=\frac{-20}{4} =-5 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+23}{2*2}=\frac{26}{4} =6+1/2 $
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