If it's not what You are looking for type in the equation solver your own equation and let us solve it.
10=19.97t-4.9t^2
We move all terms to the left:
10-(19.97t-4.9t^2)=0
We get rid of parentheses
4.9t^2-19.97t+10=0
a = 4.9; b = -19.97; c = +10;
Δ = b2-4ac
Δ = -19.972-4·4.9·10
Δ = 202.8009
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}$$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-19.97)-\sqrt{202.8009}}{2*4.9}=\frac{19.97-\sqrt{202.8009}}{9.8} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-19.97)+\sqrt{202.8009}}{2*4.9}=\frac{19.97+\sqrt{202.8009}}{9.8} $
| (14+3x)=(5x−66) | | 0=-0.06x+450,x | | 5+n/5=3 | | 1/5/x=2/0.7 | | 15/625=s^2 | | -5(r+3)=10 | | 7+9x=12x-6 | | 200=271-w | | 2x+5=10(3x-2) | | -3(x+5)+10=-16 | | 3x-2×2x-7=2×3+x-4 | | 8(m+1)+4=3(2m-8)-16 | | 1+.25x=3 | | 2(-3+x)=24 | | --3(2x-1)=30 | | 11y2+2y-1=0 | | 55=x-35 | | |5x+7|=-5 | | -3(y+5=15 | | H=-5t^2+100t+350 | | (5x-17)=(2x=50) | | -9=-t+12 | | 9+30x=11+22x | | 5x-17=2x=50 | | x²+24=60 | | 5(2x+10)=15x-10 | | −2z−(−10z+7)=2z-4-9z | | -3x+6=-2x-20 | | 2(x+5)-(4x+2x)=66 | | 2(x+5)-4x+2x)=66 | | f(5)=21(5)+14 | | 15y+5y=-15 |