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0=2y^2+11y+5
We move all terms to the left:
0-(2y^2+11y+5)=0
We add all the numbers together, and all the variables
-(2y^2+11y+5)=0
We get rid of parentheses
-2y^2-11y-5=0
a = -2; b = -11; c = -5;
Δ = b2-4ac
Δ = -112-4·(-2)·(-5)
Δ = 81
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{81}=9$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-11)-9}{2*-2}=\frac{2}{-4} =-1/2 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-11)+9}{2*-2}=\frac{20}{-4} =-5 $
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