2x^2+4x=5x+28

Solution for 2x^2+4x=5x+28 equation:

2x^2+4x=5x+28

We move all terms to the left:

2x^2+4x-(5x+28)=0

We get rid of parentheses

2x^2+4x-5x-28=0

We add all the numbers together, and all the variables

2x^2-1x-28=0

a = 2; b = -1; c = -28;
Δ = b2-4ac
Δ = -12-4·2·(-28)
Δ = 225
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{225}=15$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-15}{2*2}=\frac{-14}{4}
=-3+1/2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+15}{2*2}=\frac{16}{4}
=4
$