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2y^2-15y+22=0
a = 2; b = -15; c = +22;
Δ = b2-4ac
Δ = -152-4·2·22
Δ = 49
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{49}=7$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-15)-7}{2*2}=\frac{8}{4} =2 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-15)+7}{2*2}=\frac{22}{4} =5+1/2 $
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