0=2x^2+25x-41

Solution for 0=2x^2+25x-41 equation:

0=2x^2+25x-41

We move all terms to the left:

0-(2x^2+25x-41)=0

We add all the numbers together, and all the variables

-(2x^2+25x-41)=0

We get rid of parentheses

-2x^2-25x+41=0

a = -2; b = -25; c = +41;
Δ = b2-4ac
Δ = -252-4·(-2)·41
Δ = 953
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-25)-\sqrt{953}}{2*-2}=\frac{25-\sqrt{953}}{-4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-25)+\sqrt{953}}{2*-2}=\frac{25+\sqrt{953}}{-4}
$