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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We multiply parentheses

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


We calculate terms in parentheses: -(-(x-5)), so:

-(x-5)

We get rid of parentheses

-x+5

We add all the numbers together, and all the variables

-1x+5

Back to the equation:

-(-1x+5)


We add all the numbers together, and all the variables

x^2-1x-(-1x+5)=0

We get rid of parentheses

x^2-1x+1x-5=0

We add all the numbers together, and all the variables

x^2-5=0

a = 1; b = 0; c = -5;
Δ = b2-4ac
Δ = 02-4·1·(-5)
Δ = 20
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:

$\sqrt{\Delta}=\sqrt{20}=\sqrt{4*5}=\sqrt{4}*\sqrt{5}=2\sqrt{5}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{5}}{2*1}=\frac{0-2\sqrt{5}}{2}
=-\frac{2\sqrt{5}}{2}
=-\sqrt{5}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{5}}{2*1}=\frac{0+2\sqrt{5}}{2}
=\frac{2\sqrt{5}}{2}
=\sqrt{5}
$