(x^2+112)=(16x+131)+0

Solution for (x^2+112)=(16x+131)+0 equation:

(x^2+112)=(16x+131)+0

We move all terms to the left:

(x^2+112)-((16x+131)+0)=0

We get rid of parentheses

x^2-((16x+131)+0)+112=0


We calculate terms in parentheses: -((16x+131)+0), so:

(16x+131)+0

We add all the numbers together, and all the variables

(16x+131)

We get rid of parentheses

16x+131

Back to the equation:

-(16x+131)


We get rid of parentheses

x^2-16x-131+112=0

We add all the numbers together, and all the variables

x^2-16x-19=0

a = 1; b = -16; c = -19;
Δ = b2-4ac
Δ = -162-4·1·(-19)
Δ = 332
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:

$\sqrt{\Delta}=\sqrt{332}=\sqrt{4*83}=\sqrt{4}*\sqrt{83}=2\sqrt{83}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-2\sqrt{83}}{2*1}=\frac{16-2\sqrt{83}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+2\sqrt{83}}{2*1}=\frac{16+2\sqrt{83}}{2}
$