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0.2x^2=20
We move all terms to the left:
0.2x^2-(20)=0
a = 0.2; b = 0; c = -20;
Δ = b2-4ac
Δ = 02-4·0.2·(-20)
Δ = 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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{16}=4$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4}{2*0.2}=\frac{-4}{0.4} =-10 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4}{2*0.2}=\frac{4}{0.4} =10 $
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