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23y^2+43y=0
a = 23; b = 43; c = 0;
Δ = b2-4ac
Δ = 432-4·23·0
Δ = 1849
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{1849}=43$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(43)-43}{2*23}=\frac{-86}{46} =-1+20/23 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(43)+43}{2*23}=\frac{0}{46} =0 $
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