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0.25x^2-13x=0
a = 0.25; b = -13; c = 0;
Δ = b2-4ac
Δ = -132-4·0.25·0
Δ = 169
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{169}=13$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-13)-13}{2*0.25}=\frac{0}{0.5} =0 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-13)+13}{2*0.25}=\frac{26}{0.5} =52 $
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