What are the solutions to the equation 5t^2-8t=3 ?

The solutions to the equation 5t^2-8t=3 are t=(4+sqrt(31))/5 ,t=(4-sqrt(31))/5

Find the zeros of 5t^2-8t=3

The zeros of 5t^2-8t=3 are t=(4+sqrt(31))/5 ,t=(4-sqrt(31))/5

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5 t^{2} - 8 t = 3

t = \frac{4 + 31}{5}, t = \frac{4 - 31}{5}

Decimal Notation

t = 1.91355 \dots, t = - 0.31355 \dots

Solution steps

5 t^{2} - 8 t = 3

Move

3

to the left side

5 t^{2} - 8 t - 3 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- ( - 8 ) \pm ( - 8 ) ^{2} - 4 \cdot 5 ( - 3 )}{2 \cdot 5}

(- 8)^{2} - 4 \cdot 5 (- 3) = 231

t_{1, 2} = \frac{- ( - 8 ) \pm 2 31}{2 \cdot 5}

Separate the solutions

t_{1} = \frac{- ( - 8 ) + 2 31}{2 \cdot 5}, t_{2} = \frac{- ( - 8 ) - 2 31}{2 \cdot 5}

t = \frac{- ( - 8 ) + 2 31}{2 \cdot 5} : \frac{4 + 31}{5}

t = \frac{- ( - 8 ) - 2 31}{2 \cdot 5} : \frac{4 - 31}{5}

The solutions to the quadratic equation are:

t = \frac{4 + 31}{5}, t = \frac{4 - 31}{5}

3 x^{2} - 42 x + 144 = 0

(x + 2) (x - 8) = 0

(x - 2) (x - 2) = 8

5 2^{2} + 4 8^{2} = x^{2}

j^{2} - 4 j - 45 = 0

What are the solutions to the equation 5t^2-8t=3 ?
The solutions to the equation 5t^2-8t=3 are t=(4+sqrt(31))/5 ,t=(4-sqrt(31))/5
Find the zeros of 5t^2-8t=3
The zeros of 5t^2-8t=3 are t=(4+sqrt(31))/5 ,t=(4-sqrt(31))/5