What are the solutions to the equation 78=7t+16t^2 ?

The solutions to the equation 78=7t+16t^2 are t=2,t=-39/16

Find the zeros of 78=7t+16t^2

The zeros of 78=7t+16t^2 are t=2,t=-39/16

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78 = 7 t + 16 t^{2}

t = 2, t = - \frac{39}{16}

Decimal Notation

t = 2, t = - 2.4375

Solution steps

78 = 7 t + 16 t^{2}

Switch sides

7 t + 16 t^{2} = 78

Move

78

to the left side

7 t + 16 t^{2} - 78 = 0

Write in the standard form

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

16 t^{2} + 7 t - 78 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 7 \pm 7 ^{2} - 4 \cdot 16 ( - 78 )}{2 \cdot 16}

7^{2} - 4 \cdot 16 (- 78) = 71

t_{1, 2} = \frac{- 7 \pm 71}{2 \cdot 16}

Separate the solutions

t_{1} = \frac{- 7 + 71}{2 \cdot 16}, t_{2} = \frac{- 7 - 71}{2 \cdot 16}

t = \frac{- 7 + 71}{2 \cdot 16} : 2

t = \frac{- 7 - 71}{2 \cdot 16} : - \frac{39}{16}

The solutions to the quadratic equation are:

t = 2, t = - \frac{39}{16}

4 x^{2} + 12 x y + 9 y^{2}

- 1 \leq \frac{x + 1}{- 2} \leq 3

f = 3^{2} - 3 - 3

- \frac{5}{6} x \geq 25

16 x^{2} - 16 x - 12

What are the solutions to the equation 78=7t+16t^2 ?
The solutions to the equation 78=7t+16t^2 are t=2,t=-39/16
Find the zeros of 78=7t+16t^2
The zeros of 78=7t+16t^2 are t=2,t=-39/16