If it's not what You are looking for type in the equation solver your own equation and let us solve it.
5q^2=0
a = 5; b = 0; c = 0;
Δ = b2-4ac
Δ = 02-4·5·0
Δ = 0
Delta is equal to zero, so there is only one solution to the equation
Stosujemy wzór:$q=\frac{-b}{2a}=\frac{0}{10}=0$
| 96+4x=0+12x | | -17=-5x+8 | | 4(r+1/2)=12 | | 21+0.35p=56 | | 5k+12=13 | | k8=10 | | -2x+10=-30 | | X+13=44;x | | 3x-5x+8x=14 | | 2/3k=8+18 | | (y+34)+(y+34)=4y+8 | | 18+9x+27=90 | | 18x=10-(-8) | | 10b^2=5b | | 18+9x+27=360 | | 10^b=5b | | 4w-14w=-30 | | -35=-7/9u | | 7r^2-5r+5=5 | | 3.5(x-2)=14 | | 9.5m=-66.5 | | -24+5n=2n=24 | | 19.73+x=37.73 | | 10p−9p=13 | | 1.75p+0.25=4.25 | | 0=-3x^2+8x-4 | | 5.25x=-94.5 | | 13-45=5(x-2)-7 | | 5^(x/10)=35 | | 16x-5=3+112x | | 3g−2g=8 | | 80+.2x=30+.7x |