180=4x^2-6-2x^2+19x

Solution for 180=4x^2-6-2x^2+19x equation:

180=4x^2-6-2x^2+19x

We move all terms to the left:

180-(4x^2-6-2x^2+19x)=0

We get rid of parentheses

-4x^2+2x^2-19x+6+180=0

We add all the numbers together, and all the variables

-2x^2-19x+186=0

a = -2; b = -19; c = +186;
Δ = b2-4ac
Δ = -192-4·(-2)·186
Δ = 1849
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1849}=43$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-19)-43}{2*-2}=\frac{-24}{-4}
=+6
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-19)+43}{2*-2}=\frac{62}{-4}
=-15+1/2
$