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-3y^2+24y-45=0
a = -3; b = 24; c = -45;
Δ = b2-4ac
Δ = 242-4·(-3)·(-45)
Δ = 36
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{36}=6$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(24)-6}{2*-3}=\frac{-30}{-6} =+5 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(24)+6}{2*-3}=\frac{-18}{-6} =+3 $
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