What are the solutions to the equation 72=-16x^2+56x ?

The solutions to the equation 72=-16x^2+56x are x= 7/4-i(sqrt(23))/4 ,x= 7/4+i(sqrt(23))/4

Find the zeros of 72=-16x^2+56x

The zeros of 72=-16x^2+56x are x= 7/4-i(sqrt(23))/4 ,x= 7/4+i(sqrt(23))/4

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72 = - 16 x^{2} + 56 x

x = \frac{7}{4} - i \frac{23}{4}, x = \frac{7}{4} + i \frac{23}{4}

Solution steps

72 = - 16 x^{2} + 56 x

Switch sides

- 16 x^{2} + 56 x = 72

Move

72

to the left side

- 16 x^{2} + 56 x - 72 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 56 \pm 5 6 ^{2} - 4 ( - 16 ) ( - 72 )}{2 ( - 16 )}

Simplify

5 6^{2} - 4 (- 16) (- 72) : 823 i

x_{1, 2} = \frac{- 56 \pm 8 23 i}{2 ( - 16 )}

Separate the solutions

x_{1} = \frac{- 56 + 8 23 i}{2 ( - 16 )}, x_{2} = \frac{- 56 - 8 23 i}{2 ( - 16 )}

x = \frac{- 56 + 8 23 i}{2 ( - 16 )} : \frac{7}{4} - i \frac{23}{4}

x = \frac{- 56 - 8 23 i}{2 ( - 16 )} : \frac{7}{4} + i \frac{23}{4}

The solutions to the quadratic equation are:

x = \frac{7}{4} - i \frac{23}{4}, x = \frac{7}{4} + i \frac{23}{4}

x^{2} - 1 = 4 x

(3 x + 4)^{2} = 49

- y^{2} = 3 y - 7

18 x^{2} + 19 x - 7 = 0

3 x (x - 3) = 6 x^{2} - 11 x

What are the solutions to the equation 72=-16x^2+56x ?
The solutions to the equation 72=-16x^2+56x are x= 7/4-i(sqrt(23))/4 ,x= 7/4+i(sqrt(23))/4
Find the zeros of 72=-16x^2+56x
The zeros of 72=-16x^2+56x are x= 7/4-i(sqrt(23))/4 ,x= 7/4+i(sqrt(23))/4