2x^2+48x=104

Solution for 2x^2+48x=104 equation:

2x^2+48x=104

We move all terms to the left:

2x^2+48x-(104)=0

a = 2; b = 48; c = -104;
Δ = b2-4ac
Δ = 482-4·2·(-104)
Δ = 3136
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{3136}=56$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(48)-56}{2*2}=\frac{-104}{4}
=-26
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(48)+56}{2*2}=\frac{8}{4}
=2
$