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(2a)(a+1)=0
We multiply parentheses
2a^2+2a=0
a = 2; b = 2; c = 0;
Δ = b2-4ac
Δ = 22-4·2·0
Δ = 4
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{4}=2$$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2}{2*2}=\frac{-4}{4} =-1 $$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2}{2*2}=\frac{0}{4} =0 $
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