(x^2+12)=(43-2x)

Solution for (x^2+12)=(43-2x) equation:

(x^2+12)=(43-2x)

We move all terms to the left:

(x^2+12)-((43-2x))=0

We add all the numbers together, and all the variables

(x^2+12)-((-2x+43))=0

We get rid of parentheses

x^2-((-2x+43))+12=0


We calculate terms in parentheses: -((-2x+43)), so:

(-2x+43)

We get rid of parentheses

-2x+43

Back to the equation:

-(-2x+43)


We get rid of parentheses

x^2+2x-43+12=0

We add all the numbers together, and all the variables

x^2+2x-31=0

a = 1; b = 2; c = -31;
Δ = b2-4ac
Δ = 22-4·1·(-31)
Δ = 128
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:

$\sqrt{\Delta}=\sqrt{128}=\sqrt{64*2}=\sqrt{64}*\sqrt{2}=8\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-8\sqrt{2}}{2*1}=\frac{-2-8\sqrt{2}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+8\sqrt{2}}{2*1}=\frac{-2+8\sqrt{2}}{2}
$