What are the solutions to the equation -2x^2+16x-28=0 ?

The solutions to the equation -2x^2+16x-28=0 are x=4-sqrt(2),x=4+sqrt(2)

Find the zeros of -2x^2+16x-28=0

The zeros of -2x^2+16x-28=0 are x=4-sqrt(2),x=4+sqrt(2)

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- 2 x^{2} + 16 x - 28 = 0

x = 4 - 2, x = 4 + 2

Decimal Notation

x = 2.58578 \dots, x = 5.41421 \dots

Solution steps

- 2 x^{2} + 16 x - 28 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 16 \pm 1 6 ^{2} - 4 ( - 2 ) ( - 28 )}{2 ( - 2 )}

1 6^{2} - 4 (- 2) (- 28) = 42

x_{1, 2} = \frac{- 16 \pm 4 2}{2 ( - 2 )}

Separate the solutions

x_{1} = \frac{- 16 + 4 2}{2 ( - 2 )}, x_{2} = \frac{- 16 - 4 2}{2 ( - 2 )}

x = \frac{- 16 + 4 2}{2 ( - 2 )} : 4 - 2

x = \frac{- 16 - 4 2}{2 ( - 2 )} : 4 + 2

The solutions to the quadratic equation are:

x = 4 - 2, x = 4 + 2

(1 - 3 i)^{3}

(y + 1)^{2} = \frac{3}{2}

(2 x - 5)^{2} \geq 72 + (- 4 + x)^{2}

\frac{16}{9}

\frac{c ^{2}}{20} - \frac{c}{4} + \frac{1}{5} = 0

What are the solutions to the equation -2x^2+16x-28=0 ?
The solutions to the equation -2x^2+16x-28=0 are x=4-sqrt(2),x=4+sqrt(2)
Find the zeros of -2x^2+16x-28=0
The zeros of -2x^2+16x-28=0 are x=4-sqrt(2),x=4+sqrt(2)