32x^2+6x-4=0

Solution for 32x^2+6x-4=0 equation:

32x^2+6x-4=0

a = 32; b = 6; c = -4;
Δ = b2-4ac
Δ = 62-4·32·(-4)
Δ = 548
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{548}=\sqrt{4*137}=\sqrt{4}*\sqrt{137}=2\sqrt{137}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-2\sqrt{137}}{2*32}=\frac{-6-2\sqrt{137}}{64}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+2\sqrt{137}}{2*32}=\frac{-6+2\sqrt{137}}{64}
$