What are the solutions to the equation (6-x)(2x-5)+30=0 ?

The solutions to the equation (6-x)(2x-5)+30=0 are x=0,x= 17/2

Find the zeros of (6-x)(2x-5)+30=0

The zeros of (6-x)(2x-5)+30=0 are x=0,x= 17/2

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(6 - x) (2 x - 5) + 30 = 0

x = 0, x = \frac{17}{2}

Decimal Notation

x = 0, x = 8.5

Solution steps

(6 - x) (2 x - 5) + 30 = 0

Expand

(6 - x) (2 x - 5) + 30 : - 2 x^{2} + 17 x

- 2 x^{2} + 17 x = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 17 \pm 1 7 ^{2} - 4 ( - 2 ) \cdot 0}{2 ( - 2 )}

1 7^{2} - 4 (- 2) \cdot 0 = 17

x_{1, 2} = \frac{- 17 \pm 17}{2 ( - 2 )}

Separate the solutions

x_{1} = \frac{- 17 + 17}{2 ( - 2 )}, x_{2} = \frac{- 17 - 17}{2 ( - 2 )}

x = \frac{- 17 + 17}{2 ( - 2 )} : 0

x = \frac{- 17 - 17}{2 ( - 2 )} : \frac{17}{2}

The solutions to the quadratic equation are:

x = 0, x = \frac{17}{2}

- 0.05 x^{2} - 10 x + 480 = 0

5 - 3 x - x^{2} = 3 x^{2} + 4 x + 3

2 b^{2} + 11 b + 13 = 0

a, (a - b)^{2} = 64

x, x^{2} + y^{2} - 4 x - 10 y + 29 = 0

What are the solutions to the equation (6-x)(2x-5)+30=0 ?
The solutions to the equation (6-x)(2x-5)+30=0 are x=0,x= 17/2
Find the zeros of (6-x)(2x-5)+30=0
The zeros of (6-x)(2x-5)+30=0 are x=0,x= 17/2