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3y^2-14y+16=0
a = 3; b = -14; c = +16;
Δ = b2-4ac
Δ = -142-4·3·16
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$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-14)-2}{2*3}=\frac{12}{6} =2 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-14)+2}{2*3}=\frac{16}{6} =2+2/3 $
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