If it's not what You are looking for type in the equation solver your own equation and let us solve it.
11x^2+29x=0
a = 11; b = 29; c = 0;
Δ = b2-4ac
Δ = 292-4·11·0
Δ = 841
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{841}=29$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(29)-29}{2*11}=\frac{-58}{22} =-2+7/11 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(29)+29}{2*11}=\frac{0}{22} =0 $
| 0=100+36q-4q^2 | | F(-6)=-x^2-4x+5 | | –2d=–11.18 | | x^2+(8-x)^2=36 | | 0=q^2-9q-25 | | 2/3x+3=2/7 | | 2/3x=19/3 | | q+10=–1 | | 3=z–11 | | 3x(x-2)-8+4x=0 | | s–1=10 | | 2(x-3)+4=-5x+5 | | 3x+2/2+3×6=9 | | 3x+2/2+3.6=9 | | 1/2x-1/10=1/5x+1/2 | | 7.3y-5.18=-52.9 | | 29=2x+11 | | -3x+4=-2(-2x+5) | | X+2√x-3=0 | | X+2√x-3=O | | 5c+2+3c=-4 | | 3(-x-5)=0x+11 | | -5(4x-2)=-6-12x | | 3(-x-5)=3673.737373737383x+11 | | 3(-x-5)=1.25x+11 | | 3(-x-5)=11x+11 | | 1+6/x-27/x^2=0 | | 3(-x-5)=10x+11 | | 3(-x-5)=8x+11 | | 3(-x-5)=7x+11 | | 3(-x-5)=5x+11 | | 3x-1+x+5=112 |