2x^2-42x-125=0

Solution for 2x^2-42x-125=0 equation:

2x^2-42x-125=0

a = 2; b = -42; c = -125;
Δ = b2-4ac
Δ = -422-4·2·(-125)
Δ = 2764
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:

$\sqrt{\Delta}=\sqrt{2764}=\sqrt{4*691}=\sqrt{4}*\sqrt{691}=2\sqrt{691}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-42)-2\sqrt{691}}{2*2}=\frac{42-2\sqrt{691}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-42)+2\sqrt{691}}{2*2}=\frac{42+2\sqrt{691}}{4}
$