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x^2+5x+1.24=0
a = 1; b = 5; c = +1.24;
Δ = b2-4ac
Δ = 52-4·1·1.24
Δ = 20.04
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{20.04}}{2*1}=\frac{-5-\sqrt{20.04}}{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+\sqrt{20.04}}{2*1}=\frac{-5+\sqrt{20.04}}{2} $
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