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

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

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

We move all terms to the left:

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

-2x^2-2x=0

a = -2; b = -2; c = 0;
Δ = b2-4ac
Δ = -22-4·(-2)·0
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-2}{2*-2}=\frac{0}{-4}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2}{2*-2}=\frac{4}{-4}
=-1
$