3x^+7x^2+4x=0

Solution for 3x^+7x^2+4x=0 equation:

3x^+7x^2+4x=0

We add all the numbers together, and all the variables

7x^2+7x=0

a = 7; b = 7; c = 0;
Δ = b2-4ac
Δ = 72-4·7·0
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-7}{2*7}=\frac{-14}{14}
=-1
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+7}{2*7}=\frac{0}{14}
=0
$