What are the solutions to the equation 124=-16t^2+100t+10 ?

The solutions to the equation 124=-16t^2+100t+10 are t= 3/2 ,t= 19/4

Find the zeros of 124=-16t^2+100t+10

The zeros of 124=-16t^2+100t+10 are t= 3/2 ,t= 19/4

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124 = - 16 t^{2} + 100 t + 10

t = \frac{3}{2}, t = \frac{19}{4}

Decimal Notation

t = 1.5, t = 4.75

Solution steps

124 = - 16 t^{2} + 100 t + 10

Switch sides

- 16 t^{2} + 100 t + 10 = 124

Move

124

to the left side

- 16 t^{2} + 100 t - 114 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 100 \pm 10 0 ^{2} - 4 ( - 16 ) ( - 114 )}{2 ( - 16 )}

10 0^{2} - 4 (- 16) (- 114) = 52

t_{1, 2} = \frac{- 100 \pm 52}{2 ( - 16 )}

Separate the solutions

t_{1} = \frac{- 100 + 52}{2 ( - 16 )}, t_{2} = \frac{- 100 - 52}{2 ( - 16 )}

t = \frac{- 100 + 52}{2 ( - 16 )} : \frac{3}{2}

t = \frac{- 100 - 52}{2 ( - 16 )} : \frac{19}{4}

The solutions to the quadratic equation are:

t = \frac{3}{2}, t = \frac{19}{4}

2 x^{2} + 5 x + 2 = 0

x, y = - 48.141 x^{2} + 39.476 x

12 x^{2} - 120 x + 225 = 0

r, V = π \cdot r^{2} \cdot h

4 - x^{2} = 5

What are the solutions to the equation 124=-16t^2+100t+10 ?
The solutions to the equation 124=-16t^2+100t+10 are t= 3/2 ,t= 19/4
Find the zeros of 124=-16t^2+100t+10
The zeros of 124=-16t^2+100t+10 are t= 3/2 ,t= 19/4