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 = 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*8}=\frac{-24}{16}
=-1+1/2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+14}{2*8}=\frac{4}{16}
=1/4
$