What are the solutions to the equation (x-3)(x+7)+(x-3)(x-4)=0 ?

The solutions to the equation (x-3)(x+7)+(x-3)(x-4)=0 are x=3,x=-3/2

Find the zeros of (x-3)(x+7)+(x-3)(x-4)=0

The zeros of (x-3)(x+7)+(x-3)(x-4)=0 are x=3,x=-3/2

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(x - 3) (x + 7) + (x - 3) (x - 4) = 0

x = 3, x = - \frac{3}{2}

Decimal Notation

x = 3, x = - 1.5

Solution steps

(x - 3) (x + 7) + (x - 3) (x - 4) = 0

Expand

(x - 3) (x + 7) + (x - 3) (x - 4) : 2 x^{2} - 3 x - 9

2 x^{2} - 3 x - 9 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 \cdot 2 ( - 9 )}{2 \cdot 2}

(- 3)^{2} - 4 \cdot 2 (- 9) = 9

x_{1, 2} = \frac{- ( - 3 ) \pm 9}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- ( - 3 ) + 9}{2 \cdot 2}, x_{2} = \frac{- ( - 3 ) - 9}{2 \cdot 2}

x = \frac{- ( - 3 ) + 9}{2 \cdot 2} : 3

x = \frac{- ( - 3 ) - 9}{2 \cdot 2} : - \frac{3}{2}

The solutions to the quadratic equation are:

x = 3, x = - \frac{3}{2}

(2 x + 1)^{2} = (5 x - 1) (5 x + 1)

r^{2} = (3)^{2} + (- 1)^{2}

(4 x + 3)^{2} = 18

0.02 (0.5 x^{2} - x) = 0, 25

x, zy = x^{2}

What are the solutions to the equation (x-3)(x+7)+(x-3)(x-4)=0 ?
The solutions to the equation (x-3)(x+7)+(x-3)(x-4)=0 are x=3,x=-3/2
Find the zeros of (x-3)(x+7)+(x-3)(x-4)=0
The zeros of (x-3)(x+7)+(x-3)(x-4)=0 are x=3,x=-3/2