If it's not what You are looking for type in the equation solver your own equation and let us solve it.
5(3x)^2=0
a = 53; b = 0; c = 0;
Δ = b2-4ac
Δ = 02-4·53·0
Δ = 0
Delta is equal to zero, so there is only one solution to the equation
Stosujemy wzór:$x=\frac{-b}{2a}=\frac{0}{106}=0$
| 5(3x)^2-4(3x)=0 | | 5/5+z=3/3 | | x9-1=-5 | | 2(x+7)-6x=-2 | | (9-6)x4=12 | | (4x-14)=(2x-10) | | 2b-5=19 | | 8x-3=5x+5-7 | | 4x(3x+3)-2(3x-5)=5 | | 7x–12=16 | | 5x-2x^2+127=0 | | x-1/x+5-4=0 | | 6y–4=8y+4 | | (x-4)^2-5x-3=0 | | (x-4)2-5x-3=0 | | m+9/9=0 | | 9y=187;y= | | 38=17+x;x= | | 5w=98;w=19 | | 3x+5=13-15 | | 3x-5=11x-52 | | (3/2)x=0 | | 4x2-7x+2=0 | | X^2+36/x^2=13 | | 9x-16=4x+5 | | 22=12/x | | 0.50x+3=x | | -3x^2-30x+8=0 | | 3x^2-30x+8=0 | | 4(x-1)-(x-8)=0 | | 6a-2=5a+2 | | 6y-y2=8 |