What are the solutions to the equation solvefor t,f=(5t+2)(t-2) ?

The solutions to the equation solvefor t,f=(5t+2)(t-2) are t=(4+sqrt(5f+36))/5 ,t=(4-sqrt(5f+36))/5

Find the zeros of solvefor t,f=(5t+2)(t-2)

The zeros of solvefor t,f=(5t+2)(t-2) are t=(4+sqrt(5f+36))/5 ,t=(4-sqrt(5f+36))/5

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solve for

t, f = (5 t + 2) (t - 2)

t = \frac{4 + 5 f + 36}{5}, t = \frac{4 - 5 f + 36}{5}

Solution steps

f = (5 t + 2) (t - 2)

Expand

(5 t + 2) (t - 2) : 5 t^{2} - 8 t - 4

f = 5 t^{2} - 8 t - 4

Switch sides

5 t^{2} - 8 t - 4 = f

Move

f

to the left side

5 t^{2} - 8 t - 4 - f = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- ( - 8 ) \pm ( - 8 ) ^{2} - 4 \cdot 5 ( - 4 - f )}{2 \cdot 5}

Simplify

(- 8)^{2} - 4 \cdot 5 (- 4 - f) : 2 5 f + 36

t_{1, 2} = \frac{- ( - 8 ) \pm 2 5 f + 36}{2 \cdot 5}

Separate the solutions

t_{1} = \frac{- ( - 8 ) + 2 5 f + 36}{2 \cdot 5}, t_{2} = \frac{- ( - 8 ) - 2 5 f + 36}{2 \cdot 5}

t = \frac{- ( - 8 ) + 2 5 f + 36}{2 \cdot 5} : \frac{4 + 5 f + 36}{5}

t = \frac{- ( - 8 ) - 2 5 f + 36}{2 \cdot 5} : \frac{4 - 5 f + 36}{5}

The solutions to the quadratic equation are:

t = \frac{4 + 5 f + 36}{5}, t = \frac{4 - 5 f + 36}{5}

\frac{n ( n + 1 )}{2} = 100

121 x^{2} + 100 = 0

- 3 t^{2} + 9 t = - 810

- 8 π r^{2} + 176 π r - 768 π = 0

1 + 2 x - 3 x^{2} = 0

What are the solutions to the equation solvefor t,f=(5t+2)(t-2) ?
The solutions to the equation solvefor t,f=(5t+2)(t-2) are t=(4+sqrt(5f+36))/5 ,t=(4-sqrt(5f+36))/5
Find the zeros of solvefor t,f=(5t+2)(t-2)
The zeros of solvefor t,f=(5t+2)(t-2) are t=(4+sqrt(5f+36))/5 ,t=(4-sqrt(5f+36))/5