5=x^2+3X-4

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

5=x^2+3x-4

We move all terms to the left:

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

-1x^2-3x+9=0

a = -1; b = -3; c = +9;
Δ = b2-4ac
Δ = -32-4·(-1)·9
Δ = 45
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:

$\sqrt{\Delta}=\sqrt{45}=\sqrt{9*5}=\sqrt{9}*\sqrt{5}=3\sqrt{5}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-3\sqrt{5}}{2*-1}=\frac{3-3\sqrt{5}}{-2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+3\sqrt{5}}{2*-1}=\frac{3+3\sqrt{5}}{-2}
$