If it's not what You are looking for type in the equation solver your own equation and let us solve it.
325=c^2
We move all terms to the left:
325-(c^2)=0
We add all the numbers together, and all the variables
-1c^2+325=0
a = -1; b = 0; c = +325;
Δ = b2-4ac
Δ = 02-4·(-1)·325
Δ = 1300
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1300}=\sqrt{100*13}=\sqrt{100}*\sqrt{13}=10\sqrt{13}$$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-10\sqrt{13}}{2*-1}=\frac{0-10\sqrt{13}}{-2} =-\frac{10\sqrt{13}}{-2} =-\frac{5\sqrt{13}}{-1} $$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+10\sqrt{13}}{2*-1}=\frac{0+10\sqrt{13}}{-2} =\frac{10\sqrt{13}}{-2} =\frac{5\sqrt{13}}{-1} $
| 3b-9=(-15) | | 17=r+3 | | x^2-110=0 | | -6-4x=-58 | | x-9-5=12 | | 67=8x-93 | | 3k=310 | | 39=23.4+4.5x | | 150-1.5x=20-2.5x | | 31^2+x^2=481^2 | | (42)+(75)=(6x+15) | | 3p-2=-5 | | 39=23x+45 | | 5x-7/3=11 | | h5–5=13 | | h/5–5=13 | | 5j+1=9+4j | | a-78=99 | | 2x^2-13=20=(2x-5)(x-4) | | 21=y–9y | | y+98=121 | | 19=6f+1 | | 21=y–9y= | | 12k=240 | | -5x+9=8x-14,4 | | q+6=14q= | | -5(x+10)^7/2=-390,625 | | 3(3+5w)=5(w+1)=12 | | 8c+14=94 | | 5x/2+20=180 | | -5(x+10)^7/2=-390,635 | | 3x+33=4x=8 |