4y2+20y+25=15-2y

Solution for 4y2+20y+25=15-2y equation:

4y^2+20y+25=15-2y

We move all terms to the left:

4y^2+20y+25-(15-2y)=0

We add all the numbers together, and all the variables

4y^2+20y-(-2y+15)+25=0

We get rid of parentheses

4y^2+20y+2y-15+25=0

We add all the numbers together, and all the variables

4y^2+22y+10=0

a = 4; b = 22; c = +10;
Δ = b2-4ac
Δ = 222-4·4·10
Δ = 324
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{324}=18$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(22)-18}{2*4}=\frac{-40}{8}
=-5
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(22)+18}{2*4}=\frac{-4}{8}
=-1/2
$