X^2+13x-41=4x-5

Solution for X^2+13x-41=4x-5 equation:

X^2+13X-41=4X-5

We move all terms to the left:

X^2+13X-41-(4X-5)=0

We get rid of parentheses

X^2+13X-4X+5-41=0

We add all the numbers together, and all the variables

X^2+9X-36=0

a = 1; b = 9; c = -36;
Δ = b2-4ac
Δ = 92-4·1·(-36)
Δ = 225
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{225}=15$
$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(9)-15}{2*1}=\frac{-24}{2}
=-12
$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+15}{2*1}=\frac{6}{2}
=3
$