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0.16x^2+1.96x+1.44=0
a = 0.16; b = 1.96; c = +1.44;
Δ = b2-4ac
Δ = 1.962-4·0.16·1.44
Δ = 2.92
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{-(1.96)-\sqrt{2.92}}{2*0.16}=\frac{-1.96-\sqrt{2.92}}{0.32} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1.96)+\sqrt{2.92}}{2*0.16}=\frac{-1.96+\sqrt{2.92}}{0.32} $
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