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4y^2=-y+4
We move all terms to the left:
4y^2-(-y+4)=0
We add all the numbers together, and all the variables
4y^2-(-1y+4)=0
We get rid of parentheses
4y^2+1y-4=0
We add all the numbers together, and all the variables
4y^2+y-4=0
a = 4; b = 1; c = -4;
Δ = b2-4ac
Δ = 12-4·4·(-4)
Δ = 65
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}$$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-\sqrt{65}}{2*4}=\frac{-1-\sqrt{65}}{8} $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{65}}{2*4}=\frac{-1+\sqrt{65}}{8} $
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