42x+x^2+400=0

Solution for 42x+x^2+400=0 equation:

42x+x^2+400=0

a = 1; b = 42; c = +400;
Δ = b2-4ac
Δ = 422-4·1·400
Δ = 164
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:

$\sqrt{\Delta}=\sqrt{164}=\sqrt{4*41}=\sqrt{4}*\sqrt{41}=2\sqrt{41}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(42)-2\sqrt{41}}{2*1}=\frac{-42-2\sqrt{41}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(42)+2\sqrt{41}}{2*1}=\frac{-42+2\sqrt{41}}{2}
$