If it's not what You are looking for type in the equation solver your own equation and let us solve it.
f^2-8=0
a = 1; b = 0; c = -8;
Δ = b2-4ac
Δ = 02-4·1·(-8)
Δ = 32
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$f_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{32}=\sqrt{16*2}=\sqrt{16}*\sqrt{2}=4\sqrt{2}$$f_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{2}}{2*1}=\frac{0-4\sqrt{2}}{2} =-\frac{4\sqrt{2}}{2} =-2\sqrt{2} $$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{2}}{2*1}=\frac{0+4\sqrt{2}}{2} =\frac{4\sqrt{2}}{2} =2\sqrt{2} $
| u/4+6=6 | | f2−8=0 | | 1.3x=3.3x+2.8= | | v/16+9=10 | | 3x+2=7x= | | 3q+8=11 | | a-5/2=5+a/3 | | 6^n-1=27 | | c+13/4=4 | | 9k^2-26k+16=0 | | s5+28=108 | | 2y+16=90 | | 3x-81=2(4x-80) | | p+5/9=7 | | 6=4a=14 | | 28-(12x+164)=(100-54)+6x | | 28-(12x-164)=(100-54)-6x | | -0.6=-2(x-14) | | 28+(12x-164)=(100-54)+6x | | t/6+4=15 | | 2x-12/3=4x/5 | | 2s+18=20 | | 2x/3-4/1=4x/5 | | u/10+11=15 | | 0=5,2-x | | 10x-12x=60 | | (5x+35)+45=90 | | 8y+2=10 | | v/7+2=8 | | 4-9t=10–6t | | 3x/5-2/5=35 | | 3x-2/5=35 |