x^2+19x=7x+28

Solution for x^2+19x=7x+28 equation:

x^2+19x=7x+28

We move all terms to the left:

x^2+19x-(7x+28)=0

We get rid of parentheses

x^2+19x-7x-28=0

We add all the numbers together, and all the variables

x^2+12x-28=0

a = 1; b = 12; c = -28;
Δ = b2-4ac
Δ = 122-4·1·(-28)
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-16}{2*1}=\frac{-28}{2}
=-14
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+16}{2*1}=\frac{4}{2}
=2
$