0=y^2+6y-40

Solution for 0=y^2+6y-40 equation:

0=y^2+6y-40

We move all terms to the left:

0-(y^2+6y-40)=0

We add all the numbers together, and all the variables

-(y^2+6y-40)=0

We get rid of parentheses

-y^2-6y+40=0

We add all the numbers together, and all the variables

-1y^2-6y+40=0

a = -1; b = -6; c = +40;
Δ = b2-4ac
Δ = -62-4·(-1)·40
Δ = 196
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{196}=14$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-14}{2*-1}=\frac{-8}{-2}
=+4
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+14}{2*-1}=\frac{20}{-2}
=-10
$