If it's not what You are looking for type in the equation solver your own equation and let us solve it.
29n^2+36n=0
a = 29; b = 36; c = 0;
Δ = b2-4ac
Δ = 362-4·29·0
Δ = 1296
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{1296}=36$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(36)-36}{2*29}=\frac{-72}{58} =-1+7/29 $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(36)+36}{2*29}=\frac{0}{58} =0 $
| 12u+18=12u | | p+274=623 | | 34p2+21p=0 | | X+3+x+3+1/2x+1/2x=23 | | (3^x-2+3^x+1)=28 | | h-373=170 | | 24x+2x+6=2x+18 | | h−373=170 | | 11g=924 | | 31k2+6k=0 | | 5.6(22-3.4x=) | | z+57=97 | | n+250=953 | | h−242=–232 | | p-75=268 | | p−75=268 | | 4(w+1)=-6;w | | 4j=92 | | 697=s-29 | | 697=s−29 | | j−6=17 | | b+89=598 | | 2/5(j+40)=-4;j | | 7=j-50 | | r-479=288 | | 7p-8=22;p | | r−479=288 | | s-86=11 | | -3t+-8=25;t | | s−86=11 | | 23=g-14 | | 69=u+45 |