61x^2-32x+4=0

Solution for 61x^2-32x+4=0 equation:

61x^2-32x+4=0

a = 61; b = -32; c = +4;
Δ = b2-4ac
Δ = -322-4·61·4
Δ = 48
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:

$\sqrt{\Delta}=\sqrt{48}=\sqrt{16*3}=\sqrt{16}*\sqrt{3}=4\sqrt{3}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-32)-4\sqrt{3}}{2*61}=\frac{32-4\sqrt{3}}{122}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-32)+4\sqrt{3}}{2*61}=\frac{32+4\sqrt{3}}{122}
$