0=y^2+25y+16

Solution for 0=y^2+25y+16 equation:

0=y^2+25y+16

We move all terms to the left:

0-(y^2+25y+16)=0

We add all the numbers together, and all the variables

-(y^2+25y+16)=0

We get rid of parentheses

-y^2-25y-16=0

We add all the numbers together, and all the variables

-1y^2-25y-16=0

a = -1; b = -25; c = -16;
Δ = b2-4ac
Δ = -252-4·(-1)·(-16)
Δ = 561
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-25)-\sqrt{561}}{2*-1}=\frac{25-\sqrt{561}}{-2}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-25)+\sqrt{561}}{2*-1}=\frac{25+\sqrt{561}}{-2}
$