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3y^2-14y-5=0
a = 3; b = -14; c = -5;
Δ = b2-4ac
Δ = -142-4·3·(-5)
Δ = 256
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{256}=16$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-14)-16}{2*3}=\frac{-2}{6} =-1/3 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-14)+16}{2*3}=\frac{30}{6} =5 $
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