x^2+19x+23=8x-5

Solution for x^2+19x+23=8x-5 equation:

x^2+19x+23=8x-5

We move all terms to the left:

x^2+19x+23-(8x-5)=0

We get rid of parentheses

x^2+19x-8x+5+23=0

We add all the numbers together, and all the variables

x^2+11x+28=0

a = 1; b = 11; c = +28;
Δ = b2-4ac
Δ = 112-4·1·28
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(11)-3}{2*1}=\frac{-14}{2}
=-7
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+3}{2*1}=\frac{-8}{2}
=-4
$