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0.9=0.1x^2
We move all terms to the left:
0.9-(0.1x^2)=0
We get rid of parentheses
-0.1x^2+0.9=0
a = -0.1; b = 0; c = +0.9;
Δ = b2-4ac
Δ = 02-4·(-0.1)·0.9
Δ = 0.36
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{0.36}}{2*-0.1}=\frac{0-\sqrt{0.36}}{-0.2} =-\frac{\sqrt{}}{-0.2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+\sqrt{0.36}}{2*-0.1}=\frac{0+\sqrt{0.36}}{-0.2} =\frac{\sqrt{}}{-0.2} $
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