What are the solutions to the equation 3x=(x-4)(x-5) ?

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

Find the zeros of 3x=(x-4)(x-5)

The zeros of 3x=(x-4)(x-5) are x=10,x=2

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

x = 10, x = 2

Solution steps

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

Expand

(x - 4) (x - 5) : x^{2} - 9 x + 20

3 x = x^{2} - 9 x + 20

Switch sides

x^{2} - 9 x + 20 = 3 x

Move

3 x

to the left side

x^{2} - 12 x + 20 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 12 ) \pm ( - 12 ) ^{2} - 4 \cdot 1 \cdot 20}{2 \cdot 1}

(- 12)^{2} - 4 \cdot 1 \cdot 20 = 8

x_{1, 2} = \frac{- ( - 12 ) \pm 8}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 12 ) + 8}{2 \cdot 1}, x_{2} = \frac{- ( - 12 ) - 8}{2 \cdot 1}

x = \frac{- ( - 12 ) + 8}{2 \cdot 1} : 10

x = \frac{- ( - 12 ) - 8}{2 \cdot 1} : 2

The solutions to the quadratic equation are:

x = 10, x = 2

(t - 2)^{2} = - 16

x^{2} - 8 x + 21 = 6

3 w^{2} = 48

x (1 + x) - \frac{1}{2} (x - 2) - \frac{1}{2} x - 1 = 9

x^{2} - 42 = 0

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