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2x^2+5.3x+0.6=0
a = 2; b = 5.3; c = +0.6;
Δ = b2-4ac
Δ = 5.32-4·2·0.6
Δ = 23.29
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{-(5.3)-\sqrt{23.29}}{2*2}=\frac{-5.3-\sqrt{23.29}}{4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5.3)+\sqrt{23.29}}{2*2}=\frac{-5.3+\sqrt{23.29}}{4} $
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