What are the solutions to the equation 2d^2-d-0.36=0 ?

The solutions to the equation 2d^2-d-0.36=0 are d=(5+sqrt(97))/(20),d=(5-sqrt(97))/(20)

Find the zeros of 2d^2-d-0.36=0

The zeros of 2d^2-d-0.36=0 are d=(5+sqrt(97))/(20),d=(5-sqrt(97))/(20)

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2 d^{2} - d - 0.36 = 0

d = \frac{5 + 97}{20}, d = \frac{5 - 97}{20}

Decimal

d = 0.74244 \dots, d = - 0.24244 \dots

Solution steps

2 d^{2} - d - 0.36 = 0

Multiply both sides by

100

200 d^{2} - 100 d - 36 = 0

Solve with the quadratic formula

d_{1, 2} = \frac{- ( - 100 ) \pm ( - 100 ) ^{2} - 4 \cdot 200 ( - 36 )}{2 \cdot 200}

(- 100)^{2} - 4 \cdot 200 (- 36) = 2097

d_{1, 2} = \frac{- ( - 100 ) \pm 20 97}{2 \cdot 200}

Separate the solutions

d_{1} = \frac{- ( - 100 ) + 20 97}{2 \cdot 200}, d_{2} = \frac{- ( - 100 ) - 20 97}{2 \cdot 200}

d = \frac{- ( - 100 ) + 20 97}{2 \cdot 200} : \frac{5 + 97}{20}

d = \frac{- ( - 100 ) - 20 97}{2 \cdot 200} : \frac{5 - 97}{20}

The solutions to the quadratic equation are:

d = \frac{5 + 97}{20}, d = \frac{5 - 97}{20}

- 19 x - 19 = - 19 x + 19

19 (m + 13) < 57

(x - 7)^{2} \geq 9

(\frac{8}{17})^{4}

\frac{3 + \frac{x + 1}{y}}{6 x y}

What are the solutions to the equation 2d^2-d-0.36=0 ?
The solutions to the equation 2d^2-d-0.36=0 are d=(5+sqrt(97))/(20),d=(5-sqrt(97))/(20)
Find the zeros of 2d^2-d-0.36=0
The zeros of 2d^2-d-0.36=0 are d=(5+sqrt(97))/(20),d=(5-sqrt(97))/(20)