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50x^2+2.5x-5=0
a = 50; b = 2.5; c = -5;
Δ = b2-4ac
Δ = 2.52-4·50·(-5)
Δ = 1006.25
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2.5)-\sqrt{1006.25}}{2*50}=\frac{-2.5-\sqrt{1006.25}}{100} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2.5)+\sqrt{1006.25}}{2*50}=\frac{-2.5+\sqrt{1006.25}}{100} $
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