If it's not what You are looking for type in the equation solver your own equation and let us solve it.
4z^2+23z=0
a = 4; b = 23; c = 0;
Δ = b2-4ac
Δ = 232-4·4·0
Δ = 529
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{529}=23$$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(23)-23}{2*4}=\frac{-46}{8} =-5+3/4 $$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(23)+23}{2*4}=\frac{0}{8} =0 $
| -4-2f=10 | | 8(x+4)–6=4(x+7)+4x–2 | | 3.5d+8.25=6+4.25d | | 9x−8x=14 | | 4a-a=12 | | -2-3z=5 | | 11a-3a-7a=12 | | 5k-8k=24 | | 8s-5s=9 | | 3x=2(x-5)=50 | | 8d/25=3d | | 19c-14c=15 | | √2x-7-√x-3=√x+4 | | 6x-1=4x4x+6 | | 8m-8=16 | | 3x+9x-1=3(4x+5)= | | 14v-12v=8 | | 5(2x+3)-4(3x+5)=7 | | -2-5x= | | 16c+-15c-3c=-10 | | -18+8p=-66 | | 7h-4h=3 | | 7x+3x-6x+6=26 | | -n/3-4=2 | | 6(x-3)=4(x+7) | | 3x+9=3x-7 | | 3(y-5)+25=3y+10. | | 5(x-4)=3(x+33) | | u/11=3 | | x+172+x=180 | | 5x^2+11x-432=0 | | 20+8b=60 |