What are the solutions to the equation 17x^2-40x+8=0 ?

The solutions to the equation 17x^2-40x+8=0 are x=(2(10+sqrt(66)))/(17),x=(2(10-sqrt(66)))/(17)

Find the zeros of 17x^2-40x+8=0

The zeros of 17x^2-40x+8=0 are x=(2(10+sqrt(66)))/(17),x=(2(10-sqrt(66)))/(17)

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17 x^{2} - 40 x + 8 = 0

x = \frac{2 ( 10 + 66 )}{17}, x = \frac{2 ( 10 - 66 )}{17}

Decimal Notation

x = 2.13223 \dots, x = 0.22070 \dots

Solution steps

17 x^{2} - 40 x + 8 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 40 ) \pm ( - 40 ) ^{2} - 4 \cdot 17 \cdot 8}{2 \cdot 17}

(- 40)^{2} - 4 \cdot 17 \cdot 8 = 466

x_{1, 2} = \frac{- ( - 40 ) \pm 4 66}{2 \cdot 17}

Separate the solutions

x_{1} = \frac{- ( - 40 ) + 4 66}{2 \cdot 17}, x_{2} = \frac{- ( - 40 ) - 4 66}{2 \cdot 17}

x = \frac{- ( - 40 ) + 4 66}{2 \cdot 17} : \frac{2 ( 10 + 66 )}{17}

x = \frac{- ( - 40 ) - 4 66}{2 \cdot 17} : \frac{2 ( 10 - 66 )}{17}

The solutions to the quadratic equation are:

x = \frac{2 ( 10 + 66 )}{17}, x = \frac{2 ( 10 - 66 )}{17}

9 x^{2} + 8 (k - 5) x + 64 = 0

t, s = v_{0} \cdot t + a \cdot t^{2}

y, 9 y^{2} = x \cdot (3 - x)^{2}

y (y + 5) = 14

2 x^{2} - 5 x + (3 - k) = 0

What are the solutions to the equation 17x^2-40x+8=0 ?
The solutions to the equation 17x^2-40x+8=0 are x=(2(10+sqrt(66)))/(17),x=(2(10-sqrt(66)))/(17)
Find the zeros of 17x^2-40x+8=0
The zeros of 17x^2-40x+8=0 are x=(2(10+sqrt(66)))/(17),x=(2(10-sqrt(66)))/(17)