2,5x^2+4x-128=126

Solution for 2,5x^2+4x-128=126 equation:

2.5x^2+4x-128=126

We move all terms to the left:

2.5x^2+4x-128-(126)=0

We add all the numbers together, and all the variables

2.5x^2+4x-254=0

a = 2.5; b = 4; c = -254;
Δ = b2-4ac
Δ = 42-4·2.5·(-254)
Δ = 2556
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:

$\sqrt{\Delta}=\sqrt{2556}=\sqrt{36*71}=\sqrt{36}*\sqrt{71}=6\sqrt{71}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-6\sqrt{71}}{2*2.5}=\frac{-4-6\sqrt{71}}{5}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+6\sqrt{71}}{2*2.5}=\frac{-4+6\sqrt{71}}{5}
$