500x^2-1000x+48=0

Solution for 500x^2-1000x+48=0 equation:

500x^2-1000x+48=0

a = 500; b = -1000; c = +48;
Δ = b2-4ac
Δ = -10002-4·500·48
Δ = 904000
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:

$\sqrt{\Delta}=\sqrt{904000}=\sqrt{1600*565}=\sqrt{1600}*\sqrt{565}=40\sqrt{565}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1000)-40\sqrt{565}}{2*500}=\frac{1000-40\sqrt{565}}{1000}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1000)+40\sqrt{565}}{2*500}=\frac{1000+40\sqrt{565}}{1000}
$