What are the solutions to the equation 16x^2+51=56x ?

The solutions to the equation 16x^2+51=56x are x= 7/4+i(sqrt(2))/4 ,x= 7/4-i(sqrt(2))/4

Find the zeros of 16x^2+51=56x

The zeros of 16x^2+51=56x are x= 7/4+i(sqrt(2))/4 ,x= 7/4-i(sqrt(2))/4

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16 x^{2} + 51 = 56 x

x = \frac{7}{4} + i \frac{2}{4}, x = \frac{7}{4} - i \frac{2}{4}

Solution steps

16 x^{2} + 51 = 56 x

Move

56 x

to the left side

16 x^{2} + 51 - 56 x = 0

Write in the standard form

a x^{2} + b x + c = 0

16 x^{2} - 56 x + 51 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 56 ) \pm ( - 56 ) ^{2} - 4 \cdot 16 \cdot 51}{2 \cdot 16}

Simplify

(- 56)^{2} - 4 \cdot 16 \cdot 51 : 82 i

x_{1, 2} = \frac{- ( - 56 ) \pm 8 2 i}{2 \cdot 16}

Separate the solutions

x_{1} = \frac{- ( - 56 ) + 8 2 i}{2 \cdot 16}, x_{2} = \frac{- ( - 56 ) - 8 2 i}{2 \cdot 16}

x = \frac{- ( - 56 ) + 8 2 i}{2 \cdot 16} : \frac{7}{4} + i \frac{2}{4}

x = \frac{- ( - 56 ) - 8 2 i}{2 \cdot 16} : \frac{7}{4} - i \frac{2}{4}

The solutions to the quadratic equation are:

x = \frac{7}{4} + i \frac{2}{4}, x = \frac{7}{4} - i \frac{2}{4}

4 q (q + 2) + 10 = 0

x^{2} - 2 x - 15 = 0

9 x^{2} + 54 x = - 83

a^{2} - a - 6 = 0

- 3 + (2 x + 1)^{2} = 2 - (x + 2)^{2}

What are the solutions to the equation 16x^2+51=56x ?
The solutions to the equation 16x^2+51=56x are x= 7/4+i(sqrt(2))/4 ,x= 7/4-i(sqrt(2))/4
Find the zeros of 16x^2+51=56x
The zeros of 16x^2+51=56x are x= 7/4+i(sqrt(2))/4 ,x= 7/4-i(sqrt(2))/4