If it's not what You are looking for type in the equation solver your own equation and let us solve it.
10m^2+20m=0
a = 10; b = 20; c = 0;
Δ = b2-4ac
Δ = 202-4·10·0
Δ = 400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{400}=20$$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-20}{2*10}=\frac{-40}{20} =-2 $$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+20}{2*10}=\frac{0}{20} =0 $
| 19x−17x+3x−4x=13 | | -1=-8+b | | x2+33=0 | | x^2+21=-10x-3x2+21=−10x−3 | | -4(x-4)=92 | | 1-6x+5x^2=0 | | 5x–10=10+x | | -4+11x+85=180 | | -8x=-55 | | -2(x+1)+2x=5-2x | | 5x=105 | | 6+1x=7+1x | | (8y+42)=(5y-9) | | 180-(14x-13)=10x+15 | | 36=n4 | | 25+x+16=3x-10 | | -22=x/2 | | x^2+18x+54=4x+9x2+18x+54=4x+9 | | 5=w/3-15 | | -5x+6-9-8x=-10 | | 3/4(21c-4)=15c+15 | | (-x^2+4x+1)(x^2–8x+3)=0 | | 6x=(x-23) | | (-x2+4x+1)(x2–8x+3)=0 | | 6x+9=15+21+3x | | 25+16=3x-10 | | 3u+2u−4u−u+u=16 | | 11y-8-6y-10y=22 | | 1/5x-3/10=1/2+5/20 | | 35x+22=90 | | -23=a-10-2a | | 9j−6j+j+3j−5j=8 |