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

Solution for 4x^2+5x-4x=0 equation:

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

We add all the numbers together, and all the variables

4x^2+x=0

a = 4; b = 1; c = 0;
Δ = b2-4ac
Δ = 12-4·4·0
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

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