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-3y^2+55y-250=0
a = -3; b = 55; c = -250;
Δ = b2-4ac
Δ = 552-4·(-3)·(-250)
Δ = 25
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{25}=5$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(55)-5}{2*-3}=\frac{-60}{-6} =+10 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(55)+5}{2*-3}=\frac{-50}{-6} =8+1/3 $
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