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0=2y^2+6y
We move all terms to the left:
0-(2y^2+6y)=0
We add all the numbers together, and all the variables
-(2y^2+6y)=0
We get rid of parentheses
-2y^2-6y=0
a = -2; b = -6; c = 0;
Δ = b2-4ac
Δ = -62-4·(-2)·0
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{36}=6$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-6}{2*-2}=\frac{0}{-4} =0 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+6}{2*-2}=\frac{12}{-4} =-3 $
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