180=x+(3x+25)+(x-5)x=

Solution for 180=x+(3x+25)+(x-5)x= equation:

180=x+(3x+25)+(x-5)x=

We move all terms to the left:

180-(x+(3x+25)+(x-5)x)=0


We calculate terms in parentheses: -(x+(3x+25)+(x-5)x), so:

x+(3x+25)+(x-5)x

We multiply parentheses

x^2+x+(3x+25)-5x

We get rid of parentheses

x^2+x+3x-5x+25

We add all the numbers together, and all the variables

x^2-1x+25

Back to the equation:

-(x^2-1x+25)


We get rid of parentheses

-x^2+1x-25+180=0

We add all the numbers together, and all the variables

-1x^2+x+155=0

a = -1; b = 1; c = +155;
Δ = b2-4ac
Δ = 12-4·(-1)·155
Δ = 621
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:

$\sqrt{\Delta}=\sqrt{621}=\sqrt{9*69}=\sqrt{9}*\sqrt{69}=3\sqrt{69}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-3\sqrt{69}}{2*-1}=\frac{-1-3\sqrt{69}}{-2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+3\sqrt{69}}{2*-1}=\frac{-1+3\sqrt{69}}{-2}
$