(1-y/2)^2+y^2=12.25

Solution for (1-y/2)^2+y^2=12.25 equation:

(1-y/2)^2+y^2=12.25

We move all terms to the left:

(1-y/2)^2+y^2-(12.25)=0

We add all the numbers together, and all the variables

y^2+(-y/2+1)^2-(12.25)=0

We add all the numbers together, and all the variables

y^2+(-y/2+1)^2-12.25=0

We multiply all the terms by the denominator


y^2*2-y+1)^2+(-(12.25)*2+1)^2=0

We add all the numbers together, and all the variables

y^2*2-1y=0

Wy multiply elements

2y^2-1y=0

a = 2; b = -1; c = 0;
Δ = b2-4ac
Δ = -12-4·2·0
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-1}{2*2}=\frac{0}{4}
=0
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+1}{2*2}=\frac{2}{4}
=1/2
$