What are the solutions to the equation 16x^2-40x=-27 ?

The solutions to the equation 16x^2-40x=-27 are x= 5/4+i(sqrt(2))/4 ,x= 5/4-i(sqrt(2))/4

Find the zeros of 16x^2-40x=-27

The zeros of 16x^2-40x=-27 are x= 5/4+i(sqrt(2))/4 ,x= 5/4-i(sqrt(2))/4

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16 x^{2} - 40 x = - 27

x = \frac{5}{4} + i \frac{2}{4}, x = \frac{5}{4} - i \frac{2}{4}

Solution steps

16 x^{2} - 40 x = - 27

Move

27

to the left side

16 x^{2} - 40 x + 27 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 40 ) \pm ( - 40 ) ^{2} - 4 \cdot 16 \cdot 27}{2 \cdot 16}

Simplify

(- 40)^{2} - 4 \cdot 16 \cdot 27 : 82 i

x_{1, 2} = \frac{- ( - 40 ) \pm 8 2 i}{2 \cdot 16}

Separate the solutions

x_{1} = \frac{- ( - 40 ) + 8 2 i}{2 \cdot 16}, x_{2} = \frac{- ( - 40 ) - 8 2 i}{2 \cdot 16}

x = \frac{- ( - 40 ) + 8 2 i}{2 \cdot 16} : \frac{5}{4} + i \frac{2}{4}

x = \frac{- ( - 40 ) - 8 2 i}{2 \cdot 16} : \frac{5}{4} - i \frac{2}{4}

The solutions to the quadratic equation are:

x = \frac{5}{4} + i \frac{2}{4}, x = \frac{5}{4} - i \frac{2}{4}

x^{2} - 6 x > 7

x^{2} - ∣ 5 x + 6 ∣ = 0

((\frac{5}{2})^{2})^{3}

(x + 2)^{2} - 4 = 0

3 x^{2} + 4 x y + y^{2}

What are the solutions to the equation 16x^2-40x=-27 ?
The solutions to the equation 16x^2-40x=-27 are x= 5/4+i(sqrt(2))/4 ,x= 5/4-i(sqrt(2))/4
Find the zeros of 16x^2-40x=-27
The zeros of 16x^2-40x=-27 are x= 5/4+i(sqrt(2))/4 ,x= 5/4-i(sqrt(2))/4