64x^2+16x-3=0

Solution for 64x^2+16x-3=0 equation:

64x^2+16x-3=0

a = 64; b = 16; c = -3;
Δ = b2-4ac
Δ = 162-4·64·(-3)
Δ = 1024
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1024}=32$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-32}{2*64}=\frac{-48}{128}
=-3/8
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+32}{2*64}=\frac{16}{128}
=1/8
$