j^2-28j=-25

Solution for j^2-28j=-25 equation:

j^2-28j=-25

We move all terms to the left:

j^2-28j-(-25)=0

We add all the numbers together, and all the variables

j^2-28j+25=0

a = 1; b = -28; c = +25;
Δ = b2-4ac
Δ = -282-4·1·25
Δ = 684
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:

$\sqrt{\Delta}=\sqrt{684}=\sqrt{36*19}=\sqrt{36}*\sqrt{19}=6\sqrt{19}$
$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-28)-6\sqrt{19}}{2*1}=\frac{28-6\sqrt{19}}{2}
$
$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-28)+6\sqrt{19}}{2*1}=\frac{28+6\sqrt{19}}{2}
$