If it's not what You are looking for type in the equation solver your own equation and let us solve it.
20k^2+39k=0
a = 20; b = 39; c = 0;
Δ = b2-4ac
Δ = 392-4·20·0
Δ = 1521
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{1521}=39$$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(39)-39}{2*20}=\frac{-78}{40} =-1+19/20 $$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(39)+39}{2*20}=\frac{0}{40} =0 $
| 0.05m+20.00=37.50 | | m+20.00=37.50 | | 2(4-2r)=-2(r+5)+8r | | 47g^2-13g=0 | | 3(x+1)-2(2x-5)=6x-3 | | X=2(-3x-11)+8 | | 2m^2+20m+90=7 | | 2x+3(0)=24 | | h^2+19h-20=0 | | 36=3y-12 | | 48=3w+15 | | 5x-2+3x+70=180 | | 52=4x=12 | | (-8x+2)-(-5x+7)=0 | | q^2-24q+23=0 | | (x/4)+9=6 | | k^2+15k+14=0 | | H=-10t^2+80t+92 | | -1(1)+10=y | | s^2+48s=0 | | 4x+3(-x+10)=31 | | 7x+60=36x-56 | | x/2+12=2/3+6 | | 36+6x=9x | | -3u(u+7)=0 | | (-2x+4)-(x-11)=0 | | 3/5x+3=2x-4 | | 5v(v+9)=0 | | -x-11=2x+10= | | 2h(h+2)=0 | | 9x+6=41 | | z(3z+5)=0 |