528x+x^2=4654

Solution for 528x+x^2=4654 equation:

528x+x^2=4654

We move all terms to the left:

528x+x^2-(4654)=0

a = 1; b = 528; c = -4654;
Δ = b2-4ac
Δ = 5282-4·1·(-4654)
Δ = 297400
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:

$\sqrt{\Delta}=\sqrt{297400}=\sqrt{100*2974}=\sqrt{100}*\sqrt{2974}=10\sqrt{2974}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(528)-10\sqrt{2974}}{2*1}=\frac{-528-10\sqrt{2974}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(528)+10\sqrt{2974}}{2*1}=\frac{-528+10\sqrt{2974}}{2}
$