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x^2+x+.16=0
We add all the numbers together, and all the variables
x^2+x+0.16=0
a = 1; b = 1; c = +0.16;
Δ = b2-4ac
Δ = 12-4·1·0.16
Δ = 0.36
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{-(1)-\sqrt{0.36}}{2*1}=\frac{-1-\sqrt{0.36}}{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{0.36}}{2*1}=\frac{-1+\sqrt{0.36}}{2} $
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