x^2+12=38-16x

Solution for x^2+12=38-16x equation:

x^2+12=38-16x

We move all terms to the left:

x^2+12-(38-16x)=0

We add all the numbers together, and all the variables

x^2-(-16x+38)+12=0

We get rid of parentheses

x^2+16x-38+12=0

We add all the numbers together, and all the variables

x^2+16x-26=0

a = 1; b = 16; c = -26;
Δ = b2-4ac
Δ = 162-4·1·(-26)
Δ = 360
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:

$\sqrt{\Delta}=\sqrt{360}=\sqrt{36*10}=\sqrt{36}*\sqrt{10}=6\sqrt{10}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-6\sqrt{10}}{2*1}=\frac{-16-6\sqrt{10}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+6\sqrt{10}}{2*1}=\frac{-16+6\sqrt{10}}{2}
$