4x^2+32x-28=0

Solution for 4x^2+32x-28=0 equation:

4x^2+32x-28=0

a = 4; b = 32; c = -28;
Δ = b2-4ac
Δ = 322-4·4·(-28)
Δ = 1472
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:

$\sqrt{\Delta}=\sqrt{1472}=\sqrt{64*23}=\sqrt{64}*\sqrt{23}=8\sqrt{23}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(32)-8\sqrt{23}}{2*4}=\frac{-32-8\sqrt{23}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(32)+8\sqrt{23}}{2*4}=\frac{-32+8\sqrt{23}}{8}
$