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X^2+1.3X=0
a = 1; b = 1.3; c = 0;
Δ = b2-4ac
Δ = 1.32-4·1·0
Δ = 1.69
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{-(1.3)-\sqrt{1.69}}{2*1}=\frac{-1.3-\sqrt{1.69}}{2} $$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1.3)+\sqrt{1.69}}{2*1}=\frac{-1.3+\sqrt{1.69}}{2} $
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