If it's not what You are looking for type in the equation solver your own equation and let us solve it.
40t-4t^2=0
a = -4; b = 40; c = 0;
Δ = b2-4ac
Δ = 402-4·(-4)·0
Δ = 1600
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{1600}=40$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-40}{2*-4}=\frac{-80}{-8} =+10 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+40}{2*-4}=\frac{0}{-8} =0 $
| 180-149=y | | x=14(0.18) | | y+76=225 | | y-21=34 | | x=14(0.8) | | -3x+5=-5=-2(x+7)+2x+4 | | 2=n/10+1 | | 2(x+3)-4=24 | | 5x+7-2x+1=12-3x+1 | | 50x+150=75x | | 7x3-5x-3=-3 | | 4-12i=160 | | 2y+1=-y+3 | | 6-6y=-8-8y | | y-53=0 | | 9x-8=3x+46 | | 5x-2-2x=7 | | m=6m2m-4m=16 | | -s-1=-2s+9 | | 9x-3=84=180 | | r=10100-294r+10r | | 15x+3-4x=3(4x+1)-x | | 10f+7=237 | | 45=12w-3 | | 4-7c=11 | | x9−10=−12 | | .18x+0.9(x-5)=0.01(2x-1) | | 0.5n=12.5. | | 3x+8-x+3=39 | | 3n^2+2n=1 | | 520x11/2=x | | (x/3)+6=2 |