j^2-18j+47=0

Solution for j^2-18j+47=0 equation:

j^2-18j+47=0

a = 1; b = -18; c = +47;
Δ = b2-4ac
Δ = -182-4·1·47
Δ = 136
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{136}=\sqrt{4*34}=\sqrt{4}*\sqrt{34}=2\sqrt{34}$
$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-2\sqrt{34}}{2*1}=\frac{18-2\sqrt{34}}{2}
$
$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+2\sqrt{34}}{2*1}=\frac{18+2\sqrt{34}}{2}
$