If it's not what You are looking for type in the equation solver your own equation and let us solve it.
40v^2+19v=0
a = 40; b = 19; c = 0;
Δ = b2-4ac
Δ = 192-4·40·0
Δ = 361
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{361}=19$$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(19)-19}{2*40}=\frac{-38}{80} =-19/40 $$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+19}{2*40}=\frac{0}{80} =0 $
| (v+3)^2+7=47 | | 4(-3x+8)+12=112 | | 9^3x-1=27^x+4 | | 112=168f | | 0.5x+34=26 | | (h+8)h=16 | | 99=220d | | 2.5t=75 | | 6(x–1)–10=–52 | | X*x*3x=192 | | 16t2-64t+48=0 | | 2/3x+11=165 | | 7x+12+3x=-6(x-7)*8x | | 8n=7=7n-14 | | 3f^2-31f+10=0 | | 3y+6=54-4 | | 53+x-1/125=0 | | 60=125d | | 3-5n=-3n-1 | | 2(5y–3)=34 | | 11.21=x-3.14 | | 3n−1=8n= | | v(4)=4(40-8)(30-12) | | 9v^2+v-3=0 | | f-1(5)=3 | | 25x+250x=250 | | 3x^2-12=-20 | | 5(-6+k)-2(-3k+1)=-65 | | 3x-4(x-2)=(5+4x) | | 3x-1/2+x/3=5 | | ((x-3)+4)4=((x-6)+5)5 | | 8j-11=61 |