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2y^2-19y+45=0
a = 2; b = -19; c = +45;
Δ = b2-4ac
Δ = -192-4·2·45
Δ = 1
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{1}=1$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-19)-1}{2*2}=\frac{18}{4} =4+1/2 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-19)+1}{2*2}=\frac{20}{4} =5 $
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