x^2+16x=28

Solution for x^2+16x=28 equation:

x^2+16x=28

We move all terms to the left:

x^2+16x-(28)=0

a = 1; b = 16; c = -28;
Δ = b2-4ac
Δ = 162-4·1·(-28)
Δ = 368
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:

$\sqrt{\Delta}=\sqrt{368}=\sqrt{16*23}=\sqrt{16}*\sqrt{23}=4\sqrt{23}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-4\sqrt{23}}{2*1}=\frac{-16-4\sqrt{23}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+4\sqrt{23}}{2*1}=\frac{-16+4\sqrt{23}}{2}
$