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y^2-28y+195=0
a = 1; b = -28; c = +195;
Δ = b2-4ac
Δ = -282-4·1·195
Δ = 4
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{4}=2$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-28)-2}{2*1}=\frac{26}{2} =13 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-28)+2}{2*1}=\frac{30}{2} =15 $
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