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3+8y^2=10y
We move all terms to the left:
3+8y^2-(10y)=0
a = 8; b = -10; c = +3;
Δ = b2-4ac
Δ = -102-4·8·3
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$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-2}{2*8}=\frac{8}{16} =1/2 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+2}{2*8}=\frac{12}{16} =3/4 $
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