x^2+80x=18x

Solution for x^2+80x=18x equation:

x^2+80x=18x

We move all terms to the left:

x^2+80x-(18x)=0

We add all the numbers together, and all the variables

x^2+62x=0

a = 1; b = 62; c = 0;
Δ = b2-4ac
Δ = 622-4·1·0
Δ = 3844
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{3844}=62$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(62)-62}{2*1}=\frac{-124}{2}
=-62
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(62)+62}{2*1}=\frac{0}{2}
=0
$