What are the solutions to the equation -2(-x+5)5x+2=27 ?

The solutions to the equation -2(-x+5)5x+2=27 are x=(5+sqrt(35))/2 ,x=(5-sqrt(35))/2

Find the zeros of -2(-x+5)5x+2=27

The zeros of -2(-x+5)5x+2=27 are x=(5+sqrt(35))/2 ,x=(5-sqrt(35))/2

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- 2 (- x + 5) 5 x + 2 = 27

x = \frac{5 + 35}{2}, x = \frac{5 - 35}{2}

Decimal Notation

x = 5.45803 \dots, x = - 0.45803 \dots

Solution steps

- 2 (- x + 5) \cdot 5 x + 2 = 27

Expand

- 2 (- x + 5) \cdot 5 x + 2 : 10 x^{2} - 50 x + 2

10 x^{2} - 50 x + 2 = 27

Move

27

to the left side

10 x^{2} - 50 x - 25 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 50 ) \pm ( - 50 ) ^{2} - 4 \cdot 10 ( - 25 )}{2 \cdot 10}

(- 50)^{2} - 4 \cdot 10 (- 25) = 1035

x_{1, 2} = \frac{- ( - 50 ) \pm 10 35}{2 \cdot 10}

Separate the solutions

x_{1} = \frac{- ( - 50 ) + 10 35}{2 \cdot 10}, x_{2} = \frac{- ( - 50 ) - 10 35}{2 \cdot 10}

x = \frac{- ( - 50 ) + 10 35}{2 \cdot 10} : \frac{5 + 35}{2}

x = \frac{- ( - 50 ) - 10 35}{2 \cdot 10} : \frac{5 - 35}{2}

The solutions to the quadratic equation are:

x = \frac{5 + 35}{2}, x = \frac{5 - 35}{2}

y^{2} - 8 y - 9 = 0

(x - 4)^{2} - 3 = 0

6 x^{2} - 3 x - 3 = - 10 x

x^{2} + 2 a x = 0

5 x^{2} - \frac{3}{2} x = 0

What are the solutions to the equation -2(-x+5)5x+2=27 ?
The solutions to the equation -2(-x+5)5x+2=27 are x=(5+sqrt(35))/2 ,x=(5-sqrt(35))/2
Find the zeros of -2(-x+5)5x+2=27
The zeros of -2(-x+5)5x+2=27 are x=(5+sqrt(35))/2 ,x=(5-sqrt(35))/2