If it's not what You are looking for type in the equation solver your own equation and let us solve it.
5y^2-10y=0
a = 5; b = -10; c = 0;
Δ = b2-4ac
Δ = -102-4·5·0
Δ = 100
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{100}=10$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-10}{2*5}=\frac{0}{10} =0 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+10}{2*5}=\frac{20}{10} =2 $
| 7(x-2)=7x-2 | | 2^x+2^6-x=16 | | 8z-12=5(3z-7 | | 3(m-12)=12 | | (6x-19)=(8x+3) | | 3x+x+4=33x+16-12 | | 3x+x+4=32x+16-12 | | 3x+x+4=31x+16-12 | | 3x+x+4=30x+16-12 | | 3x+x+4=29x+16-12 | | 3x+x+4=28x+16-12 | | 139+(10x-11)+121+(9x-17)+(7x+6)+153+(9x-14)+(8x+15)=180 | | 3x+x+4=27x+16-12 | | 4x+10+x+95=180 | | 3x+x+4=26x+16-12 | | 3x+x+4=25x+16-12 | | -9z-18=-10z+20 | | 2(h+7)=18 | | 7x1+2x-1=180 | | 3(h-12)=6 | | 12z^2=z+1 | | 2v-2+2(4v+1)=-5(v+1) | | 8x+4=(x-1)+12 | | 12x-8+7=-15x+13 | | 3m=7m-8 | | 2v-2+2(4v+1)=5(v+1) | | 8x/9-2x/3=4 | | 9(7x-9)=15 | | 0.03=x+x^2 | | 7(x-7)-7=-6(-5x+8)-7x | | 3x+43=67 | | 3x+29=7(x-1) |