-28x^2-32x+48=0

Solution for -28x^2-32x+48=0 equation:

-28x^2-32x+48=0

a = -28; b = -32; c = +48;
Δ = b2-4ac
Δ = -322-4·(-28)·48
Δ = 6400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{6400}=80$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-32)-80}{2*-28}=\frac{-48}{-56}
=6/7
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-32)+80}{2*-28}=\frac{112}{-56}
=-2
$