x^2+38x=280

Solution for x^2+38x=280 equation:

x^2+38x=280

We move all terms to the left:

x^2+38x-(280)=0

a = 1; b = 38; c = -280;
Δ = b2-4ac
Δ = 382-4·1·(-280)
Δ = 2564
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:

$\sqrt{\Delta}=\sqrt{2564}=\sqrt{4*641}=\sqrt{4}*\sqrt{641}=2\sqrt{641}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(38)-2\sqrt{641}}{2*1}=\frac{-38-2\sqrt{641}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(38)+2\sqrt{641}}{2*1}=\frac{-38+2\sqrt{641}}{2}
$