If it's not what You are looking for type in the equation solver your own equation and let us solve it.
890=16t^2
We move all terms to the left:
890-(16t^2)=0
a = -16; b = 0; c = +890;
Δ = b2-4ac
Δ = 02-4·(-16)·890
Δ = 56960
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}$$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{56960}=\sqrt{64*890}=\sqrt{64}*\sqrt{890}=8\sqrt{890}$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8\sqrt{890}}{2*-16}=\frac{0-8\sqrt{890}}{-32} =-\frac{8\sqrt{890}}{-32} =-\frac{\sqrt{890}}{-4} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{890}}{2*-16}=\frac{0+8\sqrt{890}}{-32} =\frac{8\sqrt{890}}{-32} =\frac{\sqrt{890}}{-4} $
| 4(x-2)-5=6(3=x)=7 | | –2(x–5)=6(2–0.5x) | | 2(5x-10)=1250 | | 3(4)x-4=2(4) | | 7(2x-9)=-49 | | -6d+3=-3(2d-1) | | –36=6a | | 14x-56=34-x | | n2–n–182=0 | | y^2-14y=40 | | 5x-11+3x-3+3x+18=180 | | 6y+2=4y+14 | | (5x-11+3x-3+3x+18=180 | | -0.5+x=-6.75 | | 5x+30+2x=180-5x+30 | | (1/3)x-8=9 | | 7c=42= | | 10x+4+28=15x+2 | | 4x-4/5=11/6 | | d=4.9(4)^2 | | a/5.9=-17.2 | | 3(x+13)=42 | | a+1=–11 | | 211=x-(-7) | | 4c-4/5=22/5 | | 2.x+15=40 | | 2=4y=-26 | | -5-u=7 | | -2.3=-20+n | | 4.3x-2.1=6.3 | | 3x-8+9x-4=72 | | 3b^2-b-8=0 |