-2x2+19x+4x2-6=180

Solution for -2x2+19x+4x2-6=180 equation:

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

We move all terms to the left:

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