-x^2-21x+56=8+x

Solution for -x^2-21x+56=8+x equation:

-x^2-21x+56=8+x

We move all terms to the left:

-x^2-21x+56-(8+x)=0

We add all the numbers together, and all the variables

-x^2-21x-(x+8)+56=0

We add all the numbers together, and all the variables

-1x^2-21x-(x+8)+56=0

We get rid of parentheses

-1x^2-21x-x-8+56=0

We add all the numbers together, and all the variables

-1x^2-22x+48=0

a = -1; b = -22; c = +48;
Δ = b2-4ac
Δ = -222-4·(-1)·48
Δ = 676
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{676}=26$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-22)-26}{2*-1}=\frac{-4}{-2}
=+2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-22)+26}{2*-1}=\frac{48}{-2}
=-24
$