If it's not what You are looking for type in the equation solver your own equation and let us solve it.
4.9t^2+18t-16.35=0
a = 4.9; b = 18; c = -16.35;
Δ = b2-4ac
Δ = 182-4·4.9·(-16.35)
Δ = 644.46
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{-(18)-\sqrt{644.46}}{2*4.9}=\frac{-18-\sqrt{644.46}}{9.8} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+\sqrt{644.46}}{2*4.9}=\frac{-18+\sqrt{644.46}}{9.8} $
| s-1=10= | | 3x=56/2 | | /x-6=7x+1 | | 0.6+2.4=0.4x+10 | | -12(x-5)=-9(1+7x)+18 | | -c+20=-20+19c | | -9r-7=-8r | | (0.75)x=-12 | | 2v+16=6v+8 | | -7n=-4n+12 | | D-6=-t | | (-0.5)x=-14 | | -7n+3n=12-7n | | s+(−3)=10 | | 1+2x=1-3x | | 2/7=y( | | F(x)=x/26 | | x-3-3=-13 | | 9x-2-4x+3=16 | | w^2-2w=2 | | 15+x/2=11 | | -96=-6(6+2x) | | 4(2x+1)=-2(4x-5)-3x | | -0.7p=16.34+3.6p | | 5+4x=3(1-2x) | | G=-4a+1 | | 7x+4+4x+9=90° | | -3-12r=57 | | 6x-10=-x+11 | | -x+2=2x-11 | | 125=(x+15) | | 20=4d+4 |