51+20x=x^2

Solution for 51+20x=x^2 equation:

51+20x=x^2

We move all terms to the left:

51+20x-(x^2)=0

determiningTheFunctionDomain
-x^2+20x+51=0

We add all the numbers together, and all the variables

-1x^2+20x+51=0

a = -1; b = 20; c = +51;
Δ = b2-4ac
Δ = 202-4·(-1)·51
Δ = 604
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:

$\sqrt{\Delta}=\sqrt{604}=\sqrt{4*151}=\sqrt{4}*\sqrt{151}=2\sqrt{151}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-2\sqrt{151}}{2*-1}=\frac{-20-2\sqrt{151}}{-2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+2\sqrt{151}}{2*-1}=\frac{-20+2\sqrt{151}}{-2}
$