5x+x^2=398

Solution for 5x+x^2=398 equation:

5x+x^2=398

We move all terms to the left:

5x+x^2-(398)=0

a = 1; b = 5; c = -398;
Δ = b2-4ac
Δ = 52-4·1·(-398)
Δ = 1617
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:

$\sqrt{\Delta}=\sqrt{1617}=\sqrt{49*33}=\sqrt{49}*\sqrt{33}=7\sqrt{33}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-7\sqrt{33}}{2*1}=\frac{-5-7\sqrt{33}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+7\sqrt{33}}{2*1}=\frac{-5+7\sqrt{33}}{2}
$