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

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

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

We move all terms to the left:

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


We calculate terms in parentheses: -(-(x^2+3x-4)), so:

-(x^2+3x-4)

We get rid of parentheses

-x^2-3x+4

We add all the numbers together, and all the variables

-1x^2-3x+4

Back to the equation:

-(-1x^2-3x+4)


We get rid of parentheses

1x^2+3x+x-4=0

We add all the numbers together, and all the variables

x^2+4x-4=0

a = 1; b = 4; c = -4;
Δ = b2-4ac
Δ = 42-4·1·(-4)
Δ = 32
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:

$\sqrt{\Delta}=\sqrt{32}=\sqrt{16*2}=\sqrt{16}*\sqrt{2}=4\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4\sqrt{2}}{2*1}=\frac{-4-4\sqrt{2}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4\sqrt{2}}{2*1}=\frac{-4+4\sqrt{2}}{2}
$