What are the solutions to the equation 5t^2=26t+24 ?

The solutions to the equation 5t^2=26t+24 are t=6,t=-4/5

Find the zeros of 5t^2=26t+24

The zeros of 5t^2=26t+24 are t=6,t=-4/5

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5 t^{2} = 26 t + 24

t = 6, t = - \frac{4}{5}

Decimal Notation

t = 6, t = - 0.8

Solution steps

5 t^{2} = 26 t + 24

Move

24

to the left side

5 t^{2} - 24 = 26 t

Move

26 t

to the left side

5 t^{2} - 24 - 26 t = 0

Write in the standard form

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

5 t^{2} - 26 t - 24 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- ( - 26 ) \pm ( - 26 ) ^{2} - 4 \cdot 5 ( - 24 )}{2 \cdot 5}

(- 26)^{2} - 4 \cdot 5 (- 24) = 34

t_{1, 2} = \frac{- ( - 26 ) \pm 34}{2 \cdot 5}

Separate the solutions

t_{1} = \frac{- ( - 26 ) + 34}{2 \cdot 5}, t_{2} = \frac{- ( - 26 ) - 34}{2 \cdot 5}

t = \frac{- ( - 26 ) + 34}{2 \cdot 5} : 6

t = \frac{- ( - 26 ) - 34}{2 \cdot 5} : - \frac{4}{5}

The solutions to the quadratic equation are:

t = 6, t = - \frac{4}{5}

\frac{x ^{2}}{x ^{- 3}}

4 m^{2} - 5 m = 8

4 p + 1 > - 7 or 6 p + 3 < 33

2 x + 5 \geq 3 - 7 x

\frac{9}{2 45}

What are the solutions to the equation 5t^2=26t+24 ?
The solutions to the equation 5t^2=26t+24 are t=6,t=-4/5
Find the zeros of 5t^2=26t+24
The zeros of 5t^2=26t+24 are t=6,t=-4/5