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0.2x^2=-1.2x-1.6
We move all terms to the left:
0.2x^2-(-1.2x-1.6)=0
We get rid of parentheses
0.2x^2+1.2x+1.6=0
a = 0.2; b = 1.2; c = +1.6;
Δ = b2-4ac
Δ = 1.22-4·0.2·1.6
Δ = 0.16
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{-(1.2)-\sqrt{0.16}}{2*0.2}=\frac{-1.2-\sqrt{0.16}}{0.4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1.2)+\sqrt{0.16}}{2*0.2}=\frac{-1.2+\sqrt{0.16}}{0.4} $
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