2x^2-28x=-4x^2-3x-4

Solution for 2x^2-28x=-4x^2-3x-4 equation:

2x^2-28x=-4x^2-3x-4

We move all terms to the left:

2x^2-28x-(-4x^2-3x-4)=0

We get rid of parentheses

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

We add all the numbers together, and all the variables

6x^2-25x+4=0

a = 6; b = -25; c = +4;
Δ = b2-4ac
Δ = -252-4·6·4
Δ = 529
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{529}=23$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-25)-23}{2*6}=\frac{2}{12}
=1/6
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-25)+23}{2*6}=\frac{48}{12}
=4
$