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4y^2+9y=0
a = 4; b = 9; c = 0;
Δ = b2-4ac
Δ = 92-4·4·0
Δ = 81
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{81}=9$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(9)-9}{2*4}=\frac{-18}{8} =-2+1/4 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+9}{2*4}=\frac{0}{8} =0 $
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