2x^2-10*x-5=0

Solution for 2x^2-10*x-5=0 equation:

2x^2-10x-5=0

a = 2; b = -10; c = -5;
Δ = b2-4ac
Δ = -102-4·2·(-5)
Δ = 140
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:

$\sqrt{\Delta}=\sqrt{140}=\sqrt{4*35}=\sqrt{4}*\sqrt{35}=2\sqrt{35}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-2\sqrt{35}}{2*2}=\frac{10-2\sqrt{35}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+2\sqrt{35}}{2*2}=\frac{10+2\sqrt{35}}{4}
$