(4-u)(4+u)=-1

Solution for (4-u)(4+u)=-1 equation:

(4-u)(4+u)=-1

We move all terms to the left:

(4-u)(4+u)-(-1)=0

We add all the numbers together, and all the variables

(-1u+4)(u+4)-(-1)=0

We add all the numbers together, and all the variables

(-1u+4)(u+4)+1=0

We multiply parentheses ..

(-1u^2-4u+4u+16)+1=0

We get rid of parentheses

-1u^2-4u+4u+16+1=0

We add all the numbers together, and all the variables

-1u^2+17=0

a = -1; b = 0; c = +17;
Δ = b2-4ac
Δ = 02-4·(-1)·17
Δ = 68
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{68}=\sqrt{4*17}=\sqrt{4}*\sqrt{17}=2\sqrt{17}$
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{17}}{2*-1}=\frac{0-2\sqrt{17}}{-2}
=-\frac{2\sqrt{17}}{-2}
=-\frac{\sqrt{17}}{-1}
$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{17}}{2*-1}=\frac{0+2\sqrt{17}}{-2}
=\frac{2\sqrt{17}}{-2}
=\frac{\sqrt{17}}{-1}
$