What are the solutions to the equation 240x^2-160x-390=0 ?

The solutions to the equation 240x^2-160x-390=0 are x=(4+5sqrt(10))/(12),x=(4-5sqrt(10))/(12)

Find the zeros of 240x^2-160x-390=0

The zeros of 240x^2-160x-390=0 are x=(4+5sqrt(10))/(12),x=(4-5sqrt(10))/(12)

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240 x^{2} - 160 x - 390 = 0

x = \frac{4 + 5 10}{12}, x = \frac{4 - 5 10}{12}

Decimal Notation

x = 1.65094 \dots, x = - 0.98428 \dots

Solution steps

240 x^{2} - 160 x - 390 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 160 ) \pm ( - 160 ) ^{2} - 4 \cdot 240 ( - 390 )}{2 \cdot 240}

(- 160)^{2} - 4 \cdot 240 (- 390) = 20010

x_{1, 2} = \frac{- ( - 160 ) \pm 200 10}{2 \cdot 240}

Separate the solutions

x_{1} = \frac{- ( - 160 ) + 200 10}{2 \cdot 240}, x_{2} = \frac{- ( - 160 ) - 200 10}{2 \cdot 240}

x = \frac{- ( - 160 ) + 200 10}{2 \cdot 240} : \frac{4 + 5 10}{12}

x = \frac{- ( - 160 ) - 200 10}{2 \cdot 240} : \frac{4 - 5 10}{12}

The solutions to the quadratic equation are:

x = \frac{4 + 5 10}{12}, x = \frac{4 - 5 10}{12}

12 b^{2} - 83 b = 27

7 w = \frac{14}{9}

\frac{16}{3}

\frac{12 !}{9 !}

10 = - 2 + x

What are the solutions to the equation 240x^2-160x-390=0 ?
The solutions to the equation 240x^2-160x-390=0 are x=(4+5sqrt(10))/(12),x=(4-5sqrt(10))/(12)
Find the zeros of 240x^2-160x-390=0
The zeros of 240x^2-160x-390=0 are x=(4+5sqrt(10))/(12),x=(4-5sqrt(10))/(12)