-441x^2+42x+1=0

Solution for -441x^2+42x+1=0 equation:

-441x^2+42x+1=0

a = -441; b = 42; c = +1;
Δ = b2-4ac
Δ = 422-4·(-441)·1
Δ = 3528
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:

$\sqrt{\Delta}=\sqrt{3528}=\sqrt{1764*2}=\sqrt{1764}*\sqrt{2}=42\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(42)-42\sqrt{2}}{2*-441}=\frac{-42-42\sqrt{2}}{-882}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(42)+42\sqrt{2}}{2*-441}=\frac{-42+42\sqrt{2}}{-882}
$