20x^2+112x-48=20

Solution for 20x^2+112x-48=20 equation:

20x^2+112x-48=20

We move all terms to the left:

20x^2+112x-48-(20)=0

We add all the numbers together, and all the variables

20x^2+112x-68=0

a = 20; b = 112; c = -68;
Δ = b2-4ac
Δ = 1122-4·20·(-68)
Δ = 17984
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:

$\sqrt{\Delta}=\sqrt{17984}=\sqrt{64*281}=\sqrt{64}*\sqrt{281}=8\sqrt{281}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(112)-8\sqrt{281}}{2*20}=\frac{-112-8\sqrt{281}}{40}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(112)+8\sqrt{281}}{2*20}=\frac{-112+8\sqrt{281}}{40}
$