What are the solutions to the equation 12t^2-38t=15 ?

The solutions to the equation 12t^2-38t=15 are t=(19+sqrt(541))/(12),t=(19-sqrt(541))/(12)

Find the zeros of 12t^2-38t=15

The zeros of 12t^2-38t=15 are t=(19+sqrt(541))/(12),t=(19-sqrt(541))/(12)

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12 t^{2} - 38 t = 15

t = \frac{19 + 541}{12}, t = \frac{19 - 541}{12}

Decimal Notation

t = 3.52161 \dots, t = - 0.35495 \dots

Solution steps

12 t^{2} - 38 t = 15

Move

15

to the left side

12 t^{2} - 38 t - 15 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- ( - 38 ) \pm ( - 38 ) ^{2} - 4 \cdot 12 ( - 15 )}{2 \cdot 12}

(- 38)^{2} - 4 \cdot 12 (- 15) = 2541

t_{1, 2} = \frac{- ( - 38 ) \pm 2 541}{2 \cdot 12}

Separate the solutions

t_{1} = \frac{- ( - 38 ) + 2 541}{2 \cdot 12}, t_{2} = \frac{- ( - 38 ) - 2 541}{2 \cdot 12}

t = \frac{- ( - 38 ) + 2 541}{2 \cdot 12} : \frac{19 + 541}{12}

t = \frac{- ( - 38 ) - 2 541}{2 \cdot 12} : \frac{19 - 541}{12}

The solutions to the quadratic equation are:

t = \frac{19 + 541}{12}, t = \frac{19 - 541}{12}

x^{4} - (3 x)^{2} > 0

4 x = 96

- 5 x < - 35

x^{2} - 3 x + 5 \leq 0

3 - 2 (b - 2) = 2 - 7 b

What are the solutions to the equation 12t^2-38t=15 ?
The solutions to the equation 12t^2-38t=15 are t=(19+sqrt(541))/(12),t=(19-sqrt(541))/(12)
Find the zeros of 12t^2-38t=15
The zeros of 12t^2-38t=15 are t=(19+sqrt(541))/(12),t=(19-sqrt(541))/(12)