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2y^2+y+11y=0
We add all the numbers together, and all the variables
2y^2+12y=0
a = 2; b = 12; c = 0;
Δ = b2-4ac
Δ = 122-4·2·0
Δ = 144
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{144}=12$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-12}{2*2}=\frac{-24}{4} =-6 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+12}{2*2}=\frac{0}{4} =0 $
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