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0.02=10x^2
We move all terms to the left:
0.02-(10x^2)=0
a = -10; b = 0; c = +0.02;
Δ = b2-4ac
Δ = 02-4·(-10)·0.02
Δ = 0.8
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{0.8}}{2*-10}=\frac{0-\sqrt{0.8}}{-20} =-\frac{\sqrt{}}{-20} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+\sqrt{0.8}}{2*-10}=\frac{0+\sqrt{0.8}}{-20} =\frac{\sqrt{}}{-20} $
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