What are the solutions to the equation 10t^2+9t-1=6t^2 ?

The solutions to the equation 10t^2+9t-1=6t^2 are t=(-9+sqrt(97))/8 ,t=(-9-sqrt(97))/8

Find the zeros of 10t^2+9t-1=6t^2

The zeros of 10t^2+9t-1=6t^2 are t=(-9+sqrt(97))/8 ,t=(-9-sqrt(97))/8

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10 t^{2} + 9 t - 1 = 6 t^{2}

t = \frac{- 9 + 97}{8}, t = \frac{- 9 - 97}{8}

Decimal Notation

t = 0.10610 \dots, t = - 2.35610 \dots

Solution steps

10 t^{2} + 9 t - 1 = 6 t^{2}

Move

6 t^{2}

to the left side

4 t^{2} + 9 t - 1 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 9 \pm 9 ^{2} - 4 \cdot 4 ( - 1 )}{2 \cdot 4}

9^{2} - 4 \cdot 4 (- 1) = 97

t_{1, 2} = \frac{- 9 \pm 97}{2 \cdot 4}

Separate the solutions

t_{1} = \frac{- 9 + 97}{2 \cdot 4}, t_{2} = \frac{- 9 - 97}{2 \cdot 4}

t = \frac{- 9 + 97}{2 \cdot 4} : \frac{- 9 + 97}{8}

t = \frac{- 9 - 97}{2 \cdot 4} : \frac{- 9 - 97}{8}

The solutions to the quadratic equation are:

t = \frac{- 9 + 97}{8}, t = \frac{- 9 - 97}{8}

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

p^{2} + 4 p + 4 = 0

6 x^{2} + 2 x = 216

(2 x - 7) (x + 3) + 30 = x (1 - x)

64 = 2 x^{2}

What are the solutions to the equation 10t^2+9t-1=6t^2 ?
The solutions to the equation 10t^2+9t-1=6t^2 are t=(-9+sqrt(97))/8 ,t=(-9-sqrt(97))/8
Find the zeros of 10t^2+9t-1=6t^2
The zeros of 10t^2+9t-1=6t^2 are t=(-9+sqrt(97))/8 ,t=(-9-sqrt(97))/8