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 calculate terms in parentheses: -((4x^2-6)+(-2x^2+19x)), so:

(4x^2-6)+(-2x^2+19x)

determiningTheFunctionDomain
(-2x^2+19x)+(4x^2-6)

We get rid of parentheses

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

We add all the numbers together, and all the variables

2x^2+19x-6

Back to the equation:

-(2x^2+19x-6)


We get rid of parentheses

-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
$