x^2+28x-1048=0

Solution for x^2+28x-1048=0 equation:

x^2+28x-1048=0

a = 1; b = 28; c = -1048;
Δ = b2-4ac
Δ = 282-4·1·(-1048)
Δ = 4976
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:

$\sqrt{\Delta}=\sqrt{4976}=\sqrt{16*311}=\sqrt{16}*\sqrt{311}=4\sqrt{311}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(28)-4\sqrt{311}}{2*1}=\frac{-28-4\sqrt{311}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(28)+4\sqrt{311}}{2*1}=\frac{-28+4\sqrt{311}}{2}
$