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+6x+4x-8=0

We add all the numbers together, and all the variables

3x^2+10x-8=0

a = 3; b = 10; c = -8;
Δ = b2-4ac
Δ = 102-4·3·(-8)
Δ = 196
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{196}=14$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-14}{2*3}=\frac{-24}{6}
=-4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+14}{2*3}=\frac{4}{6}
=2/3
$