What are the solutions to the equation 6x^2-19x=10 ?

The solutions to the equation 6x^2-19x=10 are x=(19+sqrt(601))/(12),x=(19-sqrt(601))/(12)

Find the zeros of 6x^2-19x=10

The zeros of 6x^2-19x=10 are x=(19+sqrt(601))/(12),x=(19-sqrt(601))/(12)

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6 x^{2} - 19 x = 10

x = \frac{19 + 601}{12}, x = \frac{19 - 601}{12}

Decimal Notation

x = 3.62627 \dots, x = - 0.45960 \dots

Solution steps

6 x^{2} - 19 x = 10

Move

10

to the left side

6 x^{2} - 19 x - 10 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 19 ) \pm ( - 19 ) ^{2} - 4 \cdot 6 ( - 10 )}{2 \cdot 6}

(- 19)^{2} - 4 \cdot 6 (- 10) = 601

x_{1, 2} = \frac{- ( - 19 ) \pm 601}{2 \cdot 6}

Separate the solutions

x_{1} = \frac{- ( - 19 ) + 601}{2 \cdot 6}, x_{2} = \frac{- ( - 19 ) - 601}{2 \cdot 6}

x = \frac{- ( - 19 ) + 601}{2 \cdot 6} : \frac{19 + 601}{12}

x = \frac{- ( - 19 ) - 601}{2 \cdot 6} : \frac{19 - 601}{12}

The solutions to the quadratic equation are:

x = \frac{19 + 601}{12}, x = \frac{19 - 601}{12}

x^{2} + 7 x + 10 = 4

100000 = 350 x - 0.25 x^{2}

0 = 100 + 87 \cdot t - 5 \cdot t^{2}

- 6 x^{2} = - x - 2

(x + 2) (3 x) = (x + 2) (6)

What are the solutions to the equation 6x^2-19x=10 ?
The solutions to the equation 6x^2-19x=10 are x=(19+sqrt(601))/(12),x=(19-sqrt(601))/(12)
Find the zeros of 6x^2-19x=10
The zeros of 6x^2-19x=10 are x=(19+sqrt(601))/(12),x=(19-sqrt(601))/(12)