x(19x-18)+(7x+1)+(10x-9)=180

Solution for x(19x-18)+(7x+1)+(10x-9)=180 equation:

x(19x-18)+(7x+1)+(10x-9)=180

We move all terms to the left:

x(19x-18)+(7x+1)+(10x-9)-(180)=0

We multiply parentheses

19x^2-18x+(7x+1)+(10x-9)-180=0

We get rid of parentheses

19x^2-18x+7x+10x+1-9-180=0

We add all the numbers together, and all the variables

19x^2-1x-188=0

a = 19; b = -1; c = -188;
Δ = b2-4ac
Δ = -12-4·19·(-188)
Δ = 14289
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{-(-1)-\sqrt{14289}}{2*19}=\frac{1-\sqrt{14289}}{38}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+\sqrt{14289}}{2*19}=\frac{1+\sqrt{14289}}{38}
$