16-y2=10(4+y)

Solution for 16-y2=10(4+y) equation:

16-y2=10(4+y)

We move all terms to the left:

16-y2-(10(4+y))=0

We add all the numbers together, and all the variables

-y2-(10(y+4))+16=0

We add all the numbers together, and all the variables

-1y^2-(10(y+4))+16=0


We calculate terms in parentheses: -(10(y+4)), so:

10(y+4)

We multiply parentheses

10y+40

Back to the equation:

-(10y+40)


We get rid of parentheses

-1y^2-10y-40+16=0

We add all the numbers together, and all the variables

-1y^2-10y-24=0

a = -1; b = -10; c = -24;
Δ = b2-4ac
Δ = -102-4·(-1)·(-24)
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{-(-10)-2}{2*-1}=\frac{8}{-2}
=-4
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+2}{2*-1}=\frac{12}{-2}
=-6
$