y^2=(35+2y)

Solution for y^2=(35+2y) equation:

y^2=(35+2y)

We move all terms to the left:

y^2-((35+2y))=0

We add all the numbers together, and all the variables

y^2-((2y+35))=0


We calculate terms in parentheses: -((2y+35)), so:

(2y+35)

We get rid of parentheses

2y+35

Back to the equation:

-(2y+35)


We get rid of parentheses

y^2-2y-35=0

a = 1; b = -2; c = -35;
Δ = b2-4ac
Δ = -22-4·1·(-35)
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{144}=12$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-12}{2*1}=\frac{-10}{2}
=-5
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+12}{2*1}=\frac{14}{2}
=7
$