If it's not what You are looking for type in the equation solver your own equation and let us solve it.
20x^2+55x=0
a = 20; b = 55; c = 0;
Δ = b2-4ac
Δ = 552-4·20·0
Δ = 3025
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{3025}=55$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(55)-55}{2*20}=\frac{-110}{40} =-2+3/4 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(55)+55}{2*20}=\frac{0}{40} =0 $
| 12−4y=2y+18 | | 6=2x+4=16 | | 3(x+5)2-14(x+5)+16=0 | | 17d+3=8d+12 | | 14x-4=5x+31 | | 6u-4u-u+u=14 | | x+11+7+12/4=12 | | O4x+5=13 | | (7/4-x)-(15/7-x)=2 | | 11+7+12+x/4=12 | | 5n+2=n+22 | | 37=7-3v | | x-x/3=2(x-2) | | x-x3=2(x-2) | | 20=14+5y | | 3x2-12x+159=0 | | 10c=80c= | | 12-5y=3y-4 | | 10+x/2=12 | | 8x+2-6x=14+5x+3x | | 126=1/2b*14 | | 2.045*10^-3=(x)/(0.5-4x)^2 | | 5(s-2=3(s+4) | | 2n=n+23 | | 126=1/2*b(14) | | 1/x+1/(x+2)=28/195 | | 3k-4=-13 | | 1–(q+14)=2q+1 | | 0–(q+14)=2q+1 | | -8x+3(2x+3)=19 | | 3/4x-1=-1/6+2x | | 1/2+1/(x+2)=28/195 |