x^2-7x+41=6x+5x

Solution for x^2-7x+41=6x+5x equation:

x^2-7x+41=6x+5x

We move all terms to the left:

x^2-7x+41-(6x+5x)=0

We add all the numbers together, and all the variables

x^2-7x-(+11x)+41=0

We get rid of parentheses

x^2-7x-11x+41=0

We add all the numbers together, and all the variables

x^2-18x+41=0

a = 1; b = -18; c = +41;
Δ = b2-4ac
Δ = -182-4·1·41
Δ = 160
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:

$\sqrt{\Delta}=\sqrt{160}=\sqrt{16*10}=\sqrt{16}*\sqrt{10}=4\sqrt{10}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-4\sqrt{10}}{2*1}=\frac{18-4\sqrt{10}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+4\sqrt{10}}{2*1}=\frac{18+4\sqrt{10}}{2}
$