2x^2-48x+200=0

Solution for 2x^2-48x+200=0 equation:

2x^2-48x+200=0

a = 2; b = -48; c = +200;
Δ = b2-4ac
Δ = -482-4·2·200
Δ = 704
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:

$\sqrt{\Delta}=\sqrt{704}=\sqrt{64*11}=\sqrt{64}*\sqrt{11}=8\sqrt{11}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-48)-8\sqrt{11}}{2*2}=\frac{48-8\sqrt{11}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-48)+8\sqrt{11}}{2*2}=\frac{48+8\sqrt{11}}{4}
$