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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

4x^2+(+17x)-6-180=0

We add all the numbers together, and all the variables

4x^2+(+17x)-186=0

We get rid of parentheses

4x^2+17x-186=0

a = 4; b = 17; c = -186;
Δ = b2-4ac
Δ = 172-4·4·(-186)
Δ = 3265
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{3265}}{2*4}=\frac{-17-\sqrt{3265}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(17)+\sqrt{3265}}{2*4}=\frac{-17+\sqrt{3265}}{8}
$