2y2=(25-y)2

Solution for 2y2=(25-y)2 equation:

2y^2=(25-y)2

We move all terms to the left:

2y^2-((25-y)2)=0

We add all the numbers together, and all the variables

2y^2-((-1y+25)2)=0


We calculate terms in parentheses: -((-1y+25)2), so:

(-1y+25)2

We multiply parentheses

-2y+50

Back to the equation:

-(-2y+50)


We get rid of parentheses

2y^2+2y-50=0

a = 2; b = 2; c = -50;
Δ = b2-4ac
Δ = 22-4·2·(-50)
Δ = 404
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{404}=\sqrt{4*101}=\sqrt{4}*\sqrt{101}=2\sqrt{101}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{101}}{2*2}=\frac{-2-2\sqrt{101}}{4}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{101}}{2*2}=\frac{-2+2\sqrt{101}}{4}
$