-28x+50=x^2-31x+32

Solution for -28x+50=x^2-31x+32 equation:

-28x+50=x^2-31x+32

We move all terms to the left:

-28x+50-(x^2-31x+32)=0

We get rid of parentheses

-x^2-28x+31x-32+50=0

We add all the numbers together, and all the variables

-1x^2+3x+18=0

a = -1; b = 3; c = +18;
Δ = b2-4ac
Δ = 32-4·(-1)·18
Δ = 81
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{81}=9$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-9}{2*-1}=\frac{-12}{-2}
=+6
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+9}{2*-1}=\frac{6}{-2}
=-3
$