28x^2+200x-192=0

Solution for 28x^2+200x-192=0 equation:

28x^2+200x-192=0

a = 28; b = 200; c = -192;
Δ = b2-4ac
Δ = 2002-4·28·(-192)
Δ = 61504
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{61504}=248$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(200)-248}{2*28}=\frac{-448}{56}
=-8
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(200)+248}{2*28}=\frac{48}{56}
=6/7
$