-3=x^2+3x-4

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

-3=x^2+3x-4

We move all terms to the left:

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

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

a = -1; b = -3; c = +1;
Δ = b2-4ac
Δ = -32-4·(-1)·1
Δ = 13
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{13}}{2*-1}=\frac{3-\sqrt{13}}{-2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{13}}{2*-1}=\frac{3+\sqrt{13}}{-2}
$