64x^2+48x-7=0

Solution for 64x^2+48x-7=0 equation:

64x^2+48x-7=0

a = 64; b = 48; c = -7;
Δ = b2-4ac
Δ = 482-4·64·(-7)
Δ = 4096
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4096}=64$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(48)-64}{2*64}=\frac{-112}{128}
=-7/8
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(48)+64}{2*64}=\frac{16}{128}
=1/8
$