3x^2+6x+4x+8=

Solution for 3x^2+6x+4x+8= equation:

3x^2+6x+4x+8=

We move all terms to the left:

3x^2+6x+4x+8-()=0

We add all the numbers together, and all the variables

3x^2+10x=0

a = 3; b = 10; c = 0;
Δ = b2-4ac
Δ = 102-4·3·0
Δ = 100
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{100}=10$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-10}{2*3}=\frac{-20}{6}
=-3+1/3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+10}{2*3}=\frac{0}{6}
=0
$