3=x^2+4x-5

Solution for 3=x^2+4x-5 equation:

3=x^2+4x-5

We move all terms to the left:

3-(x^2+4x-5)=0

We get rid of parentheses

-x^2-4x+5+3=0

We add all the numbers together, and all the variables

-1x^2-4x+8=0

a = -1; b = -4; c = +8;
Δ = b2-4ac
Δ = -42-4·(-1)·8
Δ = 48
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:

$\sqrt{\Delta}=\sqrt{48}=\sqrt{16*3}=\sqrt{16}*\sqrt{3}=4\sqrt{3}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-4\sqrt{3}}{2*-1}=\frac{4-4\sqrt{3}}{-2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+4\sqrt{3}}{2*-1}=\frac{4+4\sqrt{3}}{-2}
$