1x^2+-18x+80=0

Solution for 1x^2+-18x+80=0 equation:

1x^2+-18x+80=0

We add all the numbers together, and all the variables

x^2-18x=0

a = 1; b = -18; c = 0;
Δ = b2-4ac
Δ = -182-4·1·0
Δ = 324
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{324}=18$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-18}{2*1}=\frac{0}{2}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+18}{2*1}=\frac{36}{2}
=18
$