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(x^2)+(600x)=0
a = 1; b = 600; c = 0;
Δ = b2-4ac
Δ = 6002-4·1·0
Δ = 360000
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{360000}=600$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(600)-600}{2*1}=\frac{-1200}{2} =-600 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(600)+600}{2*1}=\frac{0}{2} =0 $
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