8x^2+6x+4x+3=0

Solution for 8x^2+6x+4x+3=0 equation:

8x^2+6x+4x+3=0

We add all the numbers together, and all the variables

8x^2+10x+3=0

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

$\sqrt{\Delta}=\sqrt{4}=2$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2}{2*8}=\frac{-12}{16}
=-3/4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2}{2*8}=\frac{-8}{16}
=-1/2
$