2x2-7+4x2-6x+40+18x+3=180

Solution for 2x2-7+4x2-6x+40+18x+3=180 equation:

2x^2-7+4x^2-6x+40+18x+3=180

We move all terms to the left:

2x^2-7+4x^2-6x+40+18x+3-(180)=0

We add all the numbers together, and all the variables

6x^2+12x-144=0

a = 6; b = 12; c = -144;
Δ = b2-4ac
Δ = 122-4·6·(-144)
Δ = 3600
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{3600}=60$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-60}{2*6}=\frac{-72}{12}
=-6
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+60}{2*6}=\frac{48}{12}
=4
$