(Y+5)^2=2y^2+6y+20

Solution for (Y+5)^2=2y^2+6y+20 equation:

(+5)^2=2Y^2+6Y+20

We move all terms to the left:

(+5)^2-(2Y^2+6Y+20)=0

We add all the numbers together, and all the variables

-(2Y^2+6Y+20)+5^2=0

We add all the numbers together, and all the variables

-(2Y^2+6Y+20)+25=0

We get rid of parentheses

-2Y^2-6Y-20+25=0

We add all the numbers together, and all the variables

-2Y^2-6Y+5=0

a = -2; b = -6; c = +5;
Δ = b2-4ac
Δ = -62-4·(-2)·5
Δ = 76
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:

$\sqrt{\Delta}=\sqrt{76}=\sqrt{4*19}=\sqrt{4}*\sqrt{19}=2\sqrt{19}$
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-2\sqrt{19}}{2*-2}=\frac{6-2\sqrt{19}}{-4}
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+2\sqrt{19}}{2*-2}=\frac{6+2\sqrt{19}}{-4}
$