(X^2-4x)+(3x-2)=28

Solution for (X^2-4x)+(3x-2)=28 equation:

(X^2-4X)+(3X-2)=28

We move all terms to the left:

(X^2-4X)+(3X-2)-(28)=0

We get rid of parentheses

X^2-4X+3X-2-28=0

We add all the numbers together, and all the variables

X^2-1X-30=0

a = 1; b = -1; c = -30;
Δ = b2-4ac
Δ = -12-4·1·(-30)
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$
$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-11}{2*1}=\frac{-10}{2}
=-5
$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+11}{2*1}=\frac{12}{2}
=6
$