2x^2+x^2+11x+3x-7-5=x

Solution for 2x^2+x^2+11x+3x-7-5=x equation:

2x^2+x^2+11x+3x-7-5=x

We move all terms to the left:

2x^2+x^2+11x+3x-7-5-(x)=0

We add all the numbers together, and all the variables

3x^2+13x-12=0

a = 3; b = 13; c = -12;
Δ = b2-4ac
Δ = 132-4·3·(-12)
Δ = 313
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{-(13)-\sqrt{313}}{2*3}=\frac{-13-\sqrt{313}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+\sqrt{313}}{2*3}=\frac{-13+\sqrt{313}}{6}
$