What are the solutions to the equation 2x^2-17x+16=0 ?

The solutions to the equation 2x^2-17x+16=0 are x=(17+sqrt(161))/4 ,x=(17-sqrt(161))/4

Find the zeros of 2x^2-17x+16=0

The zeros of 2x^2-17x+16=0 are x=(17+sqrt(161))/4 ,x=(17-sqrt(161))/4

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2 x^{2} - 17 x + 16 = 0

x = \frac{17 + 161}{4}, x = \frac{17 - 161}{4}

Decimal Notation

x = 7.42214 \dots, x = 1.07785 \dots

Solution steps

2 x^{2} - 17 x + 16 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 17 ) \pm ( - 17 ) ^{2} - 4 \cdot 2 \cdot 16}{2 \cdot 2}

(- 17)^{2} - 4 \cdot 2 \cdot 16 = 161

x_{1, 2} = \frac{- ( - 17 ) \pm 161}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- ( - 17 ) + 161}{2 \cdot 2}, x_{2} = \frac{- ( - 17 ) - 161}{2 \cdot 2}

x = \frac{- ( - 17 ) + 161}{2 \cdot 2} : \frac{17 + 161}{4}

x = \frac{- ( - 17 ) - 161}{2 \cdot 2} : \frac{17 - 161}{4}

The solutions to the quadratic equation are:

x = \frac{17 + 161}{4}, x = \frac{17 - 161}{4}

(- a)^{2} - 4 (1) (- 2 a) = 0

v, v^{2} = u^{2} + 2 a s

x, (x - 0)^{2} = 4 p (y - 10)

2 y^{2} - y - 7 = 0

x, f (x + 1) = - x^{2} + x + 1

What are the solutions to the equation 2x^2-17x+16=0 ?
The solutions to the equation 2x^2-17x+16=0 are x=(17+sqrt(161))/4 ,x=(17-sqrt(161))/4
Find the zeros of 2x^2-17x+16=0
The zeros of 2x^2-17x+16=0 are x=(17+sqrt(161))/4 ,x=(17-sqrt(161))/4