0=2x^2-20x+48

Solution for 0=2x^2-20x+48 equation:

0=2x^2-20x+48

We move all terms to the left:

0-(2x^2-20x+48)=0

We add all the numbers together, and all the variables

-(2x^2-20x+48)=0

We get rid of parentheses

-2x^2+20x-48=0

a = -2; b = 20; c = -48;
Δ = b2-4ac
Δ = 202-4·(-2)·(-48)
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-4}{2*-2}=\frac{-24}{-4}
=+6
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+4}{2*-2}=\frac{-16}{-4}
=+4
$