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

The solutions to the equation (5-x)(6-x)=2 are x=7,x=4

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

The zeros of (5-x)(6-x)=2 are x=7,x=4

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(5 - x) (6 - x) = 2

x = 7, x = 4

Solution steps

(5 - x) (6 - x) = 2

Expand

(5 - x) (6 - x) : 30 - 11 x + x^{2}

30 - 11 x + x^{2} = 2

Move

2

to the left side

x^{2} - 11 x + 28 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 11 ) \pm ( - 11 ) ^{2} - 4 \cdot 1 \cdot 28}{2 \cdot 1}

(- 11)^{2} - 4 \cdot 1 \cdot 28 = 3

x_{1, 2} = \frac{- ( - 11 ) \pm 3}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 11 ) + 3}{2 \cdot 1}, x_{2} = \frac{- ( - 11 ) - 3}{2 \cdot 1}

x = \frac{- ( - 11 ) + 3}{2 \cdot 1} : 7

x = \frac{- ( - 11 ) - 3}{2 \cdot 1} : 4

The solutions to the quadratic equation are:

x = 7, x = 4

v^{2} + 2 v - 24 = 0

4 (r - 1)^{2} = 8

(5 + x) (x + 1) = 4 (3 x - 4) + (x + 1) (3 x - 4)

4 x^{2} + 36 x = - 77

4 x^{2} + 36 x = - 73

What are the solutions to the equation (5-x)(6-x)=2 ?
The solutions to the equation (5-x)(6-x)=2 are x=7,x=4
Find the zeros of (5-x)(6-x)=2
The zeros of (5-x)(6-x)=2 are x=7,x=4