2x^2+48x-2=0

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

2x^2+48x-2=0

a = 2; b = 48; c = -2;
Δ = b2-4ac
Δ = 482-4·2·(-2)
Δ = 2320
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:

$\sqrt{\Delta}=\sqrt{2320}=\sqrt{16*145}=\sqrt{16}*\sqrt{145}=4\sqrt{145}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(48)-4\sqrt{145}}{2*2}=\frac{-48-4\sqrt{145}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(48)+4\sqrt{145}}{2*2}=\frac{-48+4\sqrt{145}}{4}
$