If it's not what You are looking for type in the equation solver your own equation and let us solve it.
j^2-20j-21=0
a = 1; b = -20; c = -21;
Δ = b2-4ac
Δ = -202-4·1·(-21)
Δ = 484
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}$$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{484}=22$$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-22}{2*1}=\frac{-2}{2} =-1 $$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+22}{2*1}=\frac{42}{2} =21 $
| 1=12+-5x | | X+50x=500 | | 2/8=t/20 | | m-59=-348 | | -14s=616 | | 2√x·4x+5·2x+1+2√x=22x+2+5√x·2x+4 | | 4/3x=48/21 | | 2-9=7(2p-3) | | 3×+4y=8 | | x/29-(-864)=890 | | 60x+300/(x*x)+10x=60/66 | | j^2–20j–21=0 | | 6x+9=3x+51/5 | | 50x+x=500 | | 7=(4-5m-2) | | 4a-8=4(a-1) | | j2–20j–21=0 | | x/29-(864)=890 | | 6x+9=3x+52/5 | | 2(x=2)+3x=2(x+1)+1 | | 6x-8=11x+15 | | 4-(-7)x=13-6*x | | 5x+13=-33 | | -1t=9(1t-10)+2 | | 9x-14=2x-35 | | -5x2+7x=3 | | 2h+5=35 | | 2(z+3)+6=4(z+2)+4 | | x+(-19)=-18 | | g/29–-864=890 | | 3.5x-2=4-8.5x | | 4(t-3)+8=4(2t-5) |