(2x)^2+5x=4x+5x

Solution for (2x)^2+5x=4x+5x equation:

(2x)^2+5x=4x+5x

We move all terms to the left:

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

We add all the numbers together, and all the variables

2x^2+5x-(+9x)=0

We get rid of parentheses

2x^2+5x-9x=0

We add all the numbers together, and all the variables

2x^2-4x=0

a = 2; b = -4; c = 0;
Δ = b2-4ac
Δ = -42-4·2·0
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-4}{2*2}=\frac{0}{4}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+4}{2*2}=\frac{8}{4}
=2
$