28x+x^2=1040

Solution for 28x+x^2=1040 equation:

28x+x^2=1040

We move all terms to the left:

28x+x^2-(1040)=0

a = 1; b = 28; c = -1040;
Δ = b2-4ac
Δ = 282-4·1·(-1040)
Δ = 4944
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:

$\sqrt{\Delta}=\sqrt{4944}=\sqrt{16*309}=\sqrt{16}*\sqrt{309}=4\sqrt{309}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(28)-4\sqrt{309}}{2*1}=\frac{-28-4\sqrt{309}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(28)+4\sqrt{309}}{2*1}=\frac{-28+4\sqrt{309}}{2}
$