2y^2+14y+28=y^2+10y+25

Solution for 2y^2+14y+28=y^2+10y+25 equation:

2y^2+14y+28=y^2+10y+25

We move all terms to the left:

2y^2+14y+28-(y^2+10y+25)=0

We get rid of parentheses

2y^2-y^2+14y-10y-25+28=0

We add all the numbers together, and all the variables

y^2+4y+3=0

a = 1; b = 4; c = +3;
Δ = b2-4ac
Δ = 42-4·1·3
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-2}{2*1}=\frac{-6}{2}
=-3
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+2}{2*1}=\frac{-2}{2}
=-1
$