If it's not what You are looking for type in the equation solver your own equation and let us solve it.
16y+14y^2=0
a = 14; b = 16; c = 0;
Δ = b2-4ac
Δ = 162-4·14·0
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{256}=16$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-16}{2*14}=\frac{-32}{28} =-1+1/7 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+16}{2*14}=\frac{0}{28} =0 $
| 5x-32-8x=x+3+x | | -9b-15=5b+13 | | (X+4)(x+5)=70 | | X-20=57x-4 | | 6z+9=-6+9z | | -2.3c-6.6=12.2-3.9 | | 8u-2=-10+4u+5u | | 9x-9=7-3x | | 6(p-6)=3(p+4) | | -9-8x=-27+4x | | -3(-8-8b)=2(7b-3) | | -2x-40=81-x | | 7+5g=4g | | -1+7v=-9+6v | | -1=-3(7-4x)-4(3x+8) | | -9n+4=-10n | | 12x+0.4=25 | | -9+5b=9+8b | | 2/3-3m/8=31/24 | | 4r^2-3r-10=0 | | 4+3y=-31 | | -2k=5-k | | 5r+11=-9-5r | | 8-z=-3z | | 12+3x=4x-8 | | 8(x-5)=2(3x-10) | | 60=-4x+4(3-3x) | | 60=-4x+4(3x-3x) | | (2+3y)^4+(2-3y)^4=32 | | -4(1-3x)=4(5+x)-4x | | 67-3y=180 | | 60+1.5x=5.5x |