2(x+10)(x-4)=48

Solution for 2(x+10)(x-4)=48 equation:

2(x+10)(x-4)=48

We move all terms to the left:

2(x+10)(x-4)-(48)=0

We multiply parentheses ..

2(+x^2-4x+10x-40)-48=0

We multiply parentheses

2x^2-8x+20x-80-48=0

We add all the numbers together, and all the variables

2x^2+12x-128=0

a = 2; b = 12; c = -128;
Δ = b2-4ac
Δ = 122-4·2·(-128)
Δ = 1168
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:

$\sqrt{\Delta}=\sqrt{1168}=\sqrt{16*73}=\sqrt{16}*\sqrt{73}=4\sqrt{73}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-4\sqrt{73}}{2*2}=\frac{-12-4\sqrt{73}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+4\sqrt{73}}{2*2}=\frac{-12+4\sqrt{73}}{4}
$