2y2-11y-5=(y-5)2

Solution for 2y2-11y-5=(y-5)2 equation:

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:

(y-5)2

We multiply parentheses

2y-10

Back to the equation:

-(2y-10)


We get rid of parentheses

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}
$