-56x^2+48x=0

Solution for -56x^2+48x=0 equation:

-56x^2+48x=0

a = -56; b = 48; c = 0;
Δ = b2-4ac
Δ = 482-4·(-56)·0
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{-(48)-48}{2*-56}=\frac{-96}{-112}
=6/7
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(48)+48}{2*-56}=\frac{0}{-112}
=0
$