28x^2+20x-6=0

Solution for 28x^2+20x-6=0 equation:

28x^2+20x-6=0

a = 28; b = 20; c = -6;
Δ = b2-4ac
Δ = 202-4·28·(-6)
Δ = 1072
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{1072}=\sqrt{16*67}=\sqrt{16}*\sqrt{67}=4\sqrt{67}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-4\sqrt{67}}{2*28}=\frac{-20-4\sqrt{67}}{56}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+4\sqrt{67}}{2*28}=\frac{-20+4\sqrt{67}}{56}
$