2x^2-48x-49=0

Solution for 2x^2-48x-49=0 equation:

2x^2-48x-49=0

a = 2; b = -48; c = -49;
Δ = b2-4ac
Δ = -482-4·2·(-49)
Δ = 2696
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:

$\sqrt{\Delta}=\sqrt{2696}=\sqrt{4*674}=\sqrt{4}*\sqrt{674}=2\sqrt{674}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-48)-2\sqrt{674}}{2*2}=\frac{48-2\sqrt{674}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-48)+2\sqrt{674}}{2*2}=\frac{48+2\sqrt{674}}{4}
$