What are the solutions to the equation (x-3)(2x-1)+(x-3)(-x+5)=0 ?

The solutions to the equation (x-3)(2x-1)+(x-3)(-x+5)=0 are x=3,x=-4

Find the zeros of (x-3)(2x-1)+(x-3)(-x+5)=0

The zeros of (x-3)(2x-1)+(x-3)(-x+5)=0 are x=3,x=-4

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(x - 3) (2 x - 1) + (x - 3) (- x + 5) = 0

x = 3, x = - 4

Solution steps

(x - 3) (2 x - 1) + (x - 3) (- x + 5) = 0

Expand

(x - 3) (2 x - 1) + (x - 3) (- x + 5) : x^{2} + x - 12

x^{2} + x - 12 = 0

Solve with the quadratic formula

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

1^{2} - 4 \cdot 1 \cdot (- 12) = 7

x_{1, 2} = \frac{- 1 \pm 7}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 1 + 7}{2 \cdot 1}, x_{2} = \frac{- 1 - 7}{2 \cdot 1}

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

x = \frac{- 1 - 7}{2 \cdot 1} : - 4

The solutions to the quadratic equation are:

x = 3, x = - 4

3 \cdot (x + 2) \cdot (x - 2) = (x - 4)^{2} + 8 x

x^{2} - x + 400 = 0

(4 x + 48)^{2} = (2 x + 64)^{2}

y^{2} - 15 y + 56 = 0

- 75 x^{2} - 45 x - 6 = 0

What are the solutions to the equation (x-3)(2x-1)+(x-3)(-x+5)=0 ?
The solutions to the equation (x-3)(2x-1)+(x-3)(-x+5)=0 are x=3,x=-4
Find the zeros of (x-3)(2x-1)+(x-3)(-x+5)=0
The zeros of (x-3)(2x-1)+(x-3)(-x+5)=0 are x=3,x=-4