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2y^2=y+10
We move all terms to the left:
2y^2-(y+10)=0
We get rid of parentheses
2y^2-y-10=0
We add all the numbers together, and all the variables
2y^2-1y-10=0
a = 2; b = -1; c = -10;
Δ = b2-4ac
Δ = -12-4·2·(-10)
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$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-9}{2*2}=\frac{-8}{4} =-2 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+9}{2*2}=\frac{10}{4} =2+1/2 $
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