0=x^2+16x-40

Solution for 0=x^2+16x-40 equation:

0=x^2+16x-40

We move all terms to the left:

0-(x^2+16x-40)=0

We add all the numbers together, and all the variables

-(x^2+16x-40)=0

We get rid of parentheses

-x^2-16x+40=0

We add all the numbers together, and all the variables

-1x^2-16x+40=0

a = -1; b = -16; c = +40;
Δ = b2-4ac
Δ = -162-4·(-1)·40
Δ = 416
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{416}=\sqrt{16*26}=\sqrt{16}*\sqrt{26}=4\sqrt{26}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-4\sqrt{26}}{2*-1}=\frac{16-4\sqrt{26}}{-2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+4\sqrt{26}}{2*-1}=\frac{16+4\sqrt{26}}{-2}
$