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2=0.2x^2
We move all terms to the left:
2-(0.2x^2)=0
We get rid of parentheses
-0.2x^2+2=0
a = -0.2; b = 0; c = +2;
Δ = b2-4ac
Δ = 02-4·(-0.2)·2
Δ = 1.6
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{-(0)-\sqrt{1.6}}{2*-0.2}=\frac{0-\sqrt{1.6}}{-0.4} =-\frac{\sqrt{}}{-0.4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+\sqrt{1.6}}{2*-0.2}=\frac{0+\sqrt{1.6}}{-0.4} =\frac{\sqrt{}}{-0.4} $
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