(10x-19)(3x+4)=180

Solution for (10x-19)(3x+4)=180 equation:

(10x-19)(3x+4)=180

We move all terms to the left:

(10x-19)(3x+4)-(180)=0

We multiply parentheses ..

(+30x^2+40x-57x-76)-180=0

We get rid of parentheses

30x^2+40x-57x-76-180=0

We add all the numbers together, and all the variables

30x^2-17x-256=0

a = 30; b = -17; c = -256;
Δ = b2-4ac
Δ = -172-4·30·(-256)
Δ = 31009
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{-(-17)-\sqrt{31009}}{2*30}=\frac{17-\sqrt{31009}}{60}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-17)+\sqrt{31009}}{2*30}=\frac{17+\sqrt{31009}}{60}
$