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x2-4x+0.234=0
We add all the numbers together, and all the variables
x^2-4x+0.234=0
a = 1; b = -4; c = +0.234;
Δ = b2-4ac
Δ = -42-4·1·0.234
Δ = 15.064
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{-(-4)-\sqrt{15.064}}{2*1}=\frac{4-\sqrt{15.064}}{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+\sqrt{15.064}}{2*1}=\frac{4+\sqrt{15.064}}{2} $
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