j^2-20j=-49

Solution for j^2-20j=-49 equation:

j^2-20j=-49

We move all terms to the left:

j^2-20j-(-49)=0

We add all the numbers together, and all the variables

j^2-20j+49=0

a = 1; b = -20; c = +49;
Δ = b2-4ac
Δ = -202-4·1·49
Δ = 204
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:

$\sqrt{\Delta}=\sqrt{204}=\sqrt{4*51}=\sqrt{4}*\sqrt{51}=2\sqrt{51}$
$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-2\sqrt{51}}{2*1}=\frac{20-2\sqrt{51}}{2}
$
$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+2\sqrt{51}}{2*1}=\frac{20+2\sqrt{51}}{2}
$