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25x^2+200x=0
a = 25; b = 200; c = 0;
Δ = b2-4ac
Δ = 2002-4·25·0
Δ = 40000
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{40000}=200$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(200)-200}{2*25}=\frac{-400}{50} =-8 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(200)+200}{2*25}=\frac{0}{50} =0 $
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