0=2x^2-28x+18

Solution for 0=2x^2-28x+18 equation:

0=2x^2-28x+18

We move all terms to the left:

0-(2x^2-28x+18)=0

We add all the numbers together, and all the variables

-(2x^2-28x+18)=0

We get rid of parentheses

-2x^2+28x-18=0

a = -2; b = 28; c = -18;
Δ = b2-4ac
Δ = 282-4·(-2)·(-18)
Δ = 640
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:

$\sqrt{\Delta}=\sqrt{640}=\sqrt{64*10}=\sqrt{64}*\sqrt{10}=8\sqrt{10}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(28)-8\sqrt{10}}{2*-2}=\frac{-28-8\sqrt{10}}{-4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(28)+8\sqrt{10}}{2*-2}=\frac{-28+8\sqrt{10}}{-4}
$