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3y^2+12y-63=0
a = 3; b = 12; c = -63;
Δ = b2-4ac
Δ = 122-4·3·(-63)
Δ = 900
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{900}=30$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-30}{2*3}=\frac{-42}{6} =-7 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+30}{2*3}=\frac{18}{6} =3 $
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