X=X^2+4x+-5

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

X=X^2+4X+-5

We move all terms to the left:

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

We use the square of the difference formula

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

-1X^2-3X+5=0

a = -1; b = -3; c = +5;
Δ = b2-4ac
Δ = -32-4·(-1)·5
Δ = 29
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{-(-3)-\sqrt{29}}{2*-1}=\frac{3-\sqrt{29}}{-2}
$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{29}}{2*-1}=\frac{3+\sqrt{29}}{-2}
$