x^2+3x-41=7x

Solution for x^2+3x-41=7x equation:

x^2+3x-41=7x

We move all terms to the left:

x^2+3x-41-(7x)=0

We add all the numbers together, and all the variables

x^2-4x-41=0

a = 1; b = -4; c = -41;
Δ = b2-4ac
Δ = -42-4·1·(-41)
Δ = 180
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:

$\sqrt{\Delta}=\sqrt{180}=\sqrt{36*5}=\sqrt{36}*\sqrt{5}=6\sqrt{5}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-6\sqrt{5}}{2*1}=\frac{4-6\sqrt{5}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+6\sqrt{5}}{2*1}=\frac{4+6\sqrt{5}}{2}
$