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2y^2-11y-5=(y-5)2
We move all terms to the left:
2y^2-11y-5-((y-5)2)=0
We calculate terms in parentheses: -((y-5)2), so:We get rid of parentheses
(y-5)2
We multiply parentheses
2y-10
Back to the equation:
-(2y-10)
2y^2-11y-2y+10-5=0
We add all the numbers together, and all the variables
2y^2-13y+5=0
a = 2; b = -13; c = +5;
Δ = b2-4ac
Δ = -132-4·2·5
Δ = 129
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{-(-13)-\sqrt{129}}{2*2}=\frac{13-\sqrt{129}}{4} $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-13)+\sqrt{129}}{2*2}=\frac{13+\sqrt{129}}{4} $
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