3x+112=x^2-x

Solution for 3x+112=x^2-x equation:

3x+112=x^2-x

We move all terms to the left:

3x+112-(x^2-x)=0

We get rid of parentheses

-x^2+3x+x+112=0

We add all the numbers together, and all the variables

-1x^2+4x+112=0

a = -1; b = 4; c = +112;
Δ = b2-4ac
Δ = 42-4·(-1)·112
Δ = 464
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:

$\sqrt{\Delta}=\sqrt{464}=\sqrt{16*29}=\sqrt{16}*\sqrt{29}=4\sqrt{29}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4\sqrt{29}}{2*-1}=\frac{-4-4\sqrt{29}}{-2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4\sqrt{29}}{2*-1}=\frac{-4+4\sqrt{29}}{-2}
$