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0=2a^2+4a
We move all terms to the left:
0-(2a^2+4a)=0
We add all the numbers together, and all the variables
-(2a^2+4a)=0
We get rid of parentheses
-2a^2-4a=0
a = -2; b = -4; c = 0;
Δ = b2-4ac
Δ = -42-4·(-2)·0
Δ = 16
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{16}=4$$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-4}{2*-2}=\frac{0}{-4} =0 $$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+4}{2*-2}=\frac{8}{-4} =-2 $
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