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8y^2-28y+12=0
a = 8; b = -28; c = +12;
Δ = b2-4ac
Δ = -282-4·8·12
Δ = 400
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{400}=20$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-28)-20}{2*8}=\frac{8}{16} =1/2 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-28)+20}{2*8}=\frac{48}{16} =3 $
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