8x^2+28x^2-4x=0

Solution for 8x^2+28x^2-4x=0 equation:

8x^2+28x^2-4x=0

We add all the numbers together, and all the variables

36x^2-4x=0

a = 36; b = -4; c = 0;
Δ = b2-4ac
Δ = -42-4·36·0
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*36}=\frac{0}{72}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+4}{2*36}=\frac{8}{72}
=1/9
$