j^2-14(j)=9

Solution for j^2-14(j)=9 equation:

j^2-14(j)=9

We move all terms to the left:

j^2-14(j)-(9)=0

a = 1; b = -14; c = -9;
Δ = b2-4ac
Δ = -142-4·1·(-9)
Δ = 232
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:

$\sqrt{\Delta}=\sqrt{232}=\sqrt{4*58}=\sqrt{4}*\sqrt{58}=2\sqrt{58}$
$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-14)-2\sqrt{58}}{2*1}=\frac{14-2\sqrt{58}}{2}
$
$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-14)+2\sqrt{58}}{2*1}=\frac{14+2\sqrt{58}}{2}
$