64x^2+96x+27=0

Solution for 64x^2+96x+27=0 equation:

64x^2+96x+27=0

a = 64; b = 96; c = +27;
Δ = b2-4ac
Δ = 962-4·64·27
Δ = 2304
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2304}=48$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(96)-48}{2*64}=\frac{-144}{128}
=-1+1/8
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(96)+48}{2*64}=\frac{-48}{128}
=-3/8
$