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28x^2+51x=0
a = 28; b = 51; c = 0;
Δ = b2-4ac
Δ = 512-4·28·0
Δ = 2601
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{2601}=51$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(51)-51}{2*28}=\frac{-102}{56} =-1+23/28 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(51)+51}{2*28}=\frac{0}{56} =0 $
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