0=5x^2-10x-41

Solution for 0=5x^2-10x-41 equation:

0=5x^2-10x-41

We move all terms to the left:

0-(5x^2-10x-41)=0

We add all the numbers together, and all the variables

-(5x^2-10x-41)=0

We get rid of parentheses

-5x^2+10x+41=0

a = -5; b = 10; c = +41;
Δ = b2-4ac
Δ = 102-4·(-5)·41
Δ = 920
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:

$\sqrt{\Delta}=\sqrt{920}=\sqrt{4*230}=\sqrt{4}*\sqrt{230}=2\sqrt{230}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2\sqrt{230}}{2*-5}=\frac{-10-2\sqrt{230}}{-10}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2\sqrt{230}}{2*-5}=\frac{-10+2\sqrt{230}}{-10}
$