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1.3x^2+5x+2=0
a = 1.3; b = 5; c = +2;
Δ = b2-4ac
Δ = 52-4·1.3·2
Δ = 14.6
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{-(5)-\sqrt{14.6}}{2*1.3}=\frac{-5-\sqrt{14.6}}{2.6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+\sqrt{14.6}}{2*1.3}=\frac{-5+\sqrt{14.6}}{2.6} $
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