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0.25x^2+-0.5x=0
We add all the numbers together, and all the variables
0.25x^2-0.5x=0
a = 0.25; b = -0.5; c = 0;
Δ = b2-4ac
Δ = -0.52-4·0.25·0
Δ = 0.25
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{-(-0.5)-\sqrt{0.25}}{2*0.25}=\frac{0.5-\sqrt{0.25}}{0.5} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-0.5)+\sqrt{0.25}}{2*0.25}=\frac{0.5+\sqrt{0.25}}{0.5} $
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