If it's not what You are looking for type in the equation solver your own equation and let us solve it.
43k^2+27k=0
a = 43; b = 27; c = 0;
Δ = b2-4ac
Δ = 272-4·43·0
Δ = 729
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{729}=27$$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(27)-27}{2*43}=\frac{-54}{86} =-27/43 $$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(27)+27}{2*43}=\frac{0}{86} =0 $
| 8^(x+2)=326 | | 7e=49 | | x=0.81 | | 10=9-1/2x | | 5(x-8)+3(4-x)=32 | | 9c+4=−5c+14+13c | | y=2/3(17)-4 | | 86=6x+(-x+16) | | 86=6x+(-x-16) | | x+12=2x+48 | | 35.88=q-4.3 | | 17+6y=1 | | 4x–32=20 | | 5y+(-5)=10=1 | | 3-2k=-4 | | x*x*x*x-x*x*x-8*x*x-6x+36=0 | | 5y+(-5)=10y=1 | | 3-2k=12 | | 5y+(-5)=10-5-55y/5=5/5y | | -7(8)+y=-9 | | -7(3)+y=-9= | | 2(2x+5)+2=3(x-1)+4 | | 5y+(-5)=10-5-55y/5=5/5y=1 | | -7(0)+y=-9 | | 3(1-y)=0 | | -7(4)+y=-9 | | t+28=-7 | | 16=s+15 | | -0.1193a+5)=0.37(2a-1) | | 5x+1=-2(4x+3) | | 5x–(4–7x)=2 | | -2=2x+26 |