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x+x^2=0.3
We move all terms to the left:
x+x^2-(0.3)=0
We add all the numbers together, and all the variables
x^2+x-0.3=0
a = 1; b = 1; c = -0.3;
Δ = b2-4ac
Δ = 12-4·1·(-0.3)
Δ = 2.2
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{2.2}}{2*1}=\frac{-1-\sqrt{2.2}}{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{2.2}}{2*1}=\frac{-1+\sqrt{2.2}}{2} $
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