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-4.9a^2+a=0
a = -4.9; b = 1; c = 0;
Δ = b2-4ac
Δ = 12-4·(-4.9)·0
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{1}=1$$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-1}{2*-4.9}=\frac{-2}{-9.8} =2/9.8 $$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+1}{2*-4.9}=\frac{0}{-9.8} =0 $
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