If it's not what You are looking for type in the equation solver your own equation and let us solve it.
49k^2+29k=0
a = 49; b = 29; c = 0;
Δ = b2-4ac
Δ = 292-4·49·0
Δ = 841
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{841}=29$$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(29)-29}{2*49}=\frac{-58}{98} =-29/49 $$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(29)+29}{2*49}=\frac{0}{98} =0 $
| 15(x-5)=-95 | | 100=2/9+30x | | 100=2/9x+30 | | 100=30x+2/9 | | -85t-85t-3.9=-8.15 | | 10x-20=15x-30+5 | | 100=30=2/9x | | 11+21x=6+14x+5+7x | | 6x+9=-5x-3 | | Y/2=5x+3 | | 30=10c-10 | | 2*(3x-7=90)=180 | | 5.9-m=4.5-5m | | 3(2y+2)-y=2(y-3) | | )0.2x+0.3=0.9 | | -1(-4-x)-3=6x+36 | | 11h=10h+12 | | 70=2(4x+5)+2x | | 0.2x+0.3=0.9 | | 8h+9-9h=-h+9 | | 5(2c+1)-6(c+2=-43) | | -15=3r+1 | | 0.2x+0.3=-0.9 | | 8c=8c+2 | | 100=8-3x | | -4-0.7m=23 | | 3•m/4-m/12=7/8 | | -4x+5=3-4x+8 | | 2/2(2x+6)=3x+9 | | -3(k-3)=36 | | 3x-7=4(x-1)-x | | x+(x+35)+1.4x+(x-5)=360 |