What are the solutions to the equation 4d^2-15d+16=0 ?

The solutions to the equation 4d^2-15d+16=0 are d= 15/8+i(sqrt(31))/8 ,d= 15/8-i(sqrt(31))/8

Find the zeros of 4d^2-15d+16=0

The zeros of 4d^2-15d+16=0 are d= 15/8+i(sqrt(31))/8 ,d= 15/8-i(sqrt(31))/8

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4 d^{2} - 15 d + 16 = 0

d = \frac{15}{8} + i \frac{31}{8}, d = \frac{15}{8} - i \frac{31}{8}

Solution steps

4 d^{2} - 15 d + 16 = 0

Solve with the quadratic formula

d_{1, 2} = \frac{- ( - 15 ) \pm ( - 15 ) ^{2} - 4 \cdot 4 \cdot 16}{2 \cdot 4}

Simplify

(- 15)^{2} - 4 \cdot 4 \cdot 16 : 31 i

d_{1, 2} = \frac{- ( - 15 ) \pm 31 i}{2 \cdot 4}

Separate the solutions

d_{1} = \frac{- ( - 15 ) + 31 i}{2 \cdot 4}, d_{2} = \frac{- ( - 15 ) - 31 i}{2 \cdot 4}

d = \frac{- ( - 15 ) + 31 i}{2 \cdot 4} : \frac{15}{8} + i \frac{31}{8}

d = \frac{- ( - 15 ) - 31 i}{2 \cdot 4} : \frac{15}{8} - i \frac{31}{8}

The solutions to the quadratic equation are:

d = \frac{15}{8} + i \frac{31}{8}, d = \frac{15}{8} - i \frac{31}{8}

- 4 + ∣ 7 - 2 k ∣ < - 19

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t^{2} + 6 t + 8

1 - ln (x) \geq 0

What are the solutions to the equation 4d^2-15d+16=0 ?
The solutions to the equation 4d^2-15d+16=0 are d= 15/8+i(sqrt(31))/8 ,d= 15/8-i(sqrt(31))/8
Find the zeros of 4d^2-15d+16=0
The zeros of 4d^2-15d+16=0 are d= 15/8+i(sqrt(31))/8 ,d= 15/8-i(sqrt(31))/8