What are the solutions to the equation (10-x)(x/(4-3))=2 ?

The solutions to the equation (10-x)(x/(4-3))=2 are x=5-sqrt(23),x=5+sqrt(23)

Find the zeros of (10-x)(x/(4-3))=2

The zeros of (10-x)(x/(4-3))=2 are x=5-sqrt(23),x=5+sqrt(23)

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(10 - x) (\frac{x}{4 - 3}) = 2

x = 5 - 23, x = 5 + 23

Decimal Notation

x = 0.20416 \dots, x = 9.79583 \dots

Solution steps

(10 - x) (\frac{x}{4 - 3}) = 2

Expand

(10 - x) (\frac{x}{4 - 3}) : 10 x - x^{2}

10 x - x^{2} = 2

Move

2

to the left side

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

Write in the standard form

a x^{2} + b x + c = 0

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

Solve with the quadratic formula

x_{1, 2} = \frac{- 10 \pm 1 0 ^{2} - 4 ( - 1 ) ( - 2 )}{2 ( - 1 )}

1 0^{2} - 4 (- 1) (- 2) = 223

x_{1, 2} = \frac{- 10 \pm 2 23}{2 ( - 1 )}

Separate the solutions

x_{1} = \frac{- 10 + 2 23}{2 ( - 1 )}, x_{2} = \frac{- 10 - 2 23}{2 ( - 1 )}

x = \frac{- 10 + 2 23}{2 ( - 1 )} : 5 - 23

x = \frac{- 10 - 2 23}{2 ( - 1 )} : 5 + 23

The solutions to the quadratic equation are:

x = 5 - 23, x = 5 + 23

\frac{x}{2} = \frac{( x ^{2} )}{7}

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

\frac{( x ^{2} + 8 )}{2} = 3 x

6 x^{2} - 17 = 7

x, x^{2} + y^{2} + z^{2} - 2 x - 4 y - 6 z + 1 = 0

What are the solutions to the equation (10-x)(x/(4-3))=2 ?
The solutions to the equation (10-x)(x/(4-3))=2 are x=5-sqrt(23),x=5+sqrt(23)
Find the zeros of (10-x)(x/(4-3))=2
The zeros of (10-x)(x/(4-3))=2 are x=5-sqrt(23),x=5+sqrt(23)