-x^2+3x=x-19

Solution for -x^2+3x=x-19 equation:

-x^2+3x=x-19

We move all terms to the left:

-x^2+3x-(x-19)=0

We add all the numbers together, and all the variables

-1x^2+3x-(x-19)=0

We get rid of parentheses

-1x^2+3x-x+19=0

We add all the numbers together, and all the variables

-1x^2+2x+19=0

a = -1; b = 2; c = +19;
Δ = b2-4ac
Δ = 22-4·(-1)·19
Δ = 80
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:

$\sqrt{\Delta}=\sqrt{80}=\sqrt{16*5}=\sqrt{16}*\sqrt{5}=4\sqrt{5}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-4\sqrt{5}}{2*-1}=\frac{-2-4\sqrt{5}}{-2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+4\sqrt{5}}{2*-1}=\frac{-2+4\sqrt{5}}{-2}
$