3x^2+-6x+18=0

Solution for 3x^2+-6x+18=0 equation:

3x^2+-6x+18=0

We add all the numbers together, and all the variables

3x^2-6x=0

a = 3; b = -6; c = 0;
Δ = b2-4ac
Δ = -62-4·3·0
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-6}{2*3}=\frac{0}{6}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+6}{2*3}=\frac{12}{6}
=2
$