128=10x+x^2

Solution for 128=10x+x^2 equation:

128=10x+x^2

We move all terms to the left:

128-(10x+x^2)=0

We get rid of parentheses

-x^2-10x+128=0

We add all the numbers together, and all the variables

-1x^2-10x+128=0

a = -1; b = -10; c = +128;
Δ = b2-4ac
Δ = -102-4·(-1)·128
Δ = 612
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:

$\sqrt{\Delta}=\sqrt{612}=\sqrt{36*17}=\sqrt{36}*\sqrt{17}=6\sqrt{17}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-6\sqrt{17}}{2*-1}=\frac{10-6\sqrt{17}}{-2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+6\sqrt{17}}{2*-1}=\frac{10+6\sqrt{17}}{-2}
$