2x+(x)^2=12

Solution for 2x+(x)^2=12 equation:

2x+(x)^2=12

We move all terms to the left:

2x+(x)^2-(12)=0

determiningTheFunctionDomain
x^2+2x-12=0

a = 1; b = 2; c = -12;
Δ = b2-4ac
Δ = 22-4·1·(-12)
Δ = 52
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{52}=\sqrt{4*13}=\sqrt{4}*\sqrt{13}=2\sqrt{13}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{13}}{2*1}=\frac{-2-2\sqrt{13}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{13}}{2*1}=\frac{-2+2\sqrt{13}}{2}
$