4x(5x+40)=0

Solution for 4x(5x+40)=0 equation:

4x(5x+40)=0

We multiply parentheses

20x^2+160x=0

a = 20; b = 160; c = 0;
Δ = b2-4ac
Δ = 1602-4·20·0
Δ = 25600
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25600}=160$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(160)-160}{2*20}=\frac{-320}{40}
=-8
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(160)+160}{2*20}=\frac{0}{40}
=0
$