x^2+(-9x)+18=0

Solution for x^2+(-9x)+18=0 equation:

x^2+(-9x)+18=0

We get rid of parentheses

x^2-9x+18=0

a = 1; b = -9; c = +18;
Δ = b2-4ac
Δ = -92-4·1·18
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-9)-3}{2*1}=\frac{6}{2}
=3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+3}{2*1}=\frac{12}{2}
=6
$