What are the solutions to the equation 3t^2-5t=1 ?

The solutions to the equation 3t^2-5t=1 are t=(5+sqrt(37))/6 ,t=(5-sqrt(37))/6

Find the zeros of 3t^2-5t=1

The zeros of 3t^2-5t=1 are t=(5+sqrt(37))/6 ,t=(5-sqrt(37))/6

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3 t^{2} - 5 t = 1

t = \frac{5 + 37}{6}, t = \frac{5 - 37}{6}

Decimal Notation

t = 1.84712 \dots, t = - 0.18046 \dots

Solution steps

3 t^{2} - 5 t = 1

Move

1

to the left side

3 t^{2} - 5 t - 1 = 0

Solve with the quadratic formula

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

(- 5)^{2} - 4 \cdot 3 (- 1) = 37

t_{1, 2} = \frac{- ( - 5 ) \pm 37}{2 \cdot 3}

Separate the solutions

t_{1} = \frac{- ( - 5 ) + 37}{2 \cdot 3}, t_{2} = \frac{- ( - 5 ) - 37}{2 \cdot 3}

t = \frac{- ( - 5 ) + 37}{2 \cdot 3} : \frac{5 + 37}{6}

t = \frac{- ( - 5 ) - 37}{2 \cdot 3} : \frac{5 - 37}{6}

The solutions to the quadratic equation are:

t = \frac{5 + 37}{6}, t = \frac{5 - 37}{6}

5 x^{2} = 3 x - 4

x^{2} + (- \frac{3}{6})^{2} = 1

5 x^{2} = 3 x + 8

t^{2} + \frac{11}{3} t + 1 = 0

3 x^{2} - x = x^{2} - 4 x

What are the solutions to the equation 3t^2-5t=1 ?
The solutions to the equation 3t^2-5t=1 are t=(5+sqrt(37))/6 ,t=(5-sqrt(37))/6
Find the zeros of 3t^2-5t=1
The zeros of 3t^2-5t=1 are t=(5+sqrt(37))/6 ,t=(5-sqrt(37))/6