2x^2-32x-169=0

Solution for 2x^2-32x-169=0 equation:

2x^2-32x-169=0

a = 2; b = -32; c = -169;
Δ = b2-4ac
Δ = -322-4·2·(-169)
Δ = 2376
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:

$\sqrt{\Delta}=\sqrt{2376}=\sqrt{36*66}=\sqrt{36}*\sqrt{66}=6\sqrt{66}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-32)-6\sqrt{66}}{2*2}=\frac{32-6\sqrt{66}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-32)+6\sqrt{66}}{2*2}=\frac{32+6\sqrt{66}}{4}
$