What are the solutions to the equation 18=2t+16t^2 ?

The solutions to the equation 18=2t+16t^2 are t=1,t=-9/8

Find the zeros of 18=2t+16t^2

The zeros of 18=2t+16t^2 are t=1,t=-9/8

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18 = 2 t + 16 t^{2}

t = 1, t = - \frac{9}{8}

Solution steps

18 = 2 t + 16 t^{2}

Switch sides

2 t + 16 t^{2} = 18

Move

18

to the left side

2 t + 16 t^{2} - 18 = 0

Write in the standard form

a x^{2} + b x + c = 0

16 t^{2} + 2 t - 18 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 \cdot 16 ( - 18 )}{2 \cdot 16}

2^{2} - 4 \cdot 16 (- 18) = 34

t_{1, 2} = \frac{- 2 \pm 34}{2 \cdot 16}

Separate the solutions

t_{1} = \frac{- 2 + 34}{2 \cdot 16}, t_{2} = \frac{- 2 - 34}{2 \cdot 16}

t = \frac{- 2 + 34}{2 \cdot 16} : 1

t = \frac{- 2 - 34}{2 \cdot 16} : - \frac{9}{8}

The solutions to the quadratic equation are:

t = 1, t = - \frac{9}{8}

- \frac{21}{7}

x^{2} - 2 x \leq 2

3 x^{2} - x - 3 = 0

\frac{3}{4} \cdot 16

∣ x - 7 ∣ = - 1

What are the solutions to the equation 18=2t+16t^2 ?
The solutions to the equation 18=2t+16t^2 are t=1,t=-9/8
Find the zeros of 18=2t+16t^2
The zeros of 18=2t+16t^2 are t=1,t=-9/8