If it's not what You are looking for type in the equation solver your own equation and let us solve it.
14z^2+42z=0
a = 14; b = 42; c = 0;
Δ = b2-4ac
Δ = 422-4·14·0
Δ = 1764
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{1764}=42$$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(42)-42}{2*14}=\frac{-84}{28} =-3 $$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(42)+42}{2*14}=\frac{0}{28} =0 $
| 15w=32+7w | | 125+x+133=180 | | .6667x+3=11 | | -190+40y=6 | | 42-6u=u | | -83=-3x-4(x+12) | | 20x=123 | | -190+40y+9y=6 | | 0.5(x-12)+2=1.25(x-8)-9.5 | | 50=8y-30 | | 75+18x+3=180 | | n+8)+(n−12)= | | 3(y)=15(y-2) | | 2y-20+2y-20+y+y=720 | | 10(-19+4y)+9y=6 | | -144=-2(8x+6)-6x | | 18x+3=105 | | .98=1x | | 32=-15+x | | -9u+19.27=-16.66-12.2u+4.89 | | 5w-2=5w+10 | | 1.96=w5 | | x=-4/3+8 | | 2x1=-15 | | 20h-6=3h+19h+20 | | y=-38-2 | | 082x+0810=0 | | 10+6u=16 | | 14j-20=6-12j | | 35x=191 | | 1.96=5x | | 4x+11=3x+8 |