What are the solutions to the equation x(x-7)+12=0 ?

The solutions to the equation x(x-7)+12=0 are x=4,x=3

Find the zeros of x(x-7)+12=0

The zeros of x(x-7)+12=0 are x=4,x=3

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x (x - 7) + 12 = 0

x = 4, x = 3

Solution steps

x (x - 7) + 12 = 0

Expand

x (x - 7) + 12 : x^{2} - 7 x + 12

x^{2} - 7 x + 12 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 7 ) \pm ( - 7 ) ^{2} - 4 \cdot 1 \cdot 12}{2 \cdot 1}

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

x = 4, x = 3

2 (x + 5)^{2} - 32 = 0

2 x^{2} - (k - 1) x - k = 0

3 y^{2} = 81

x, 4 x^{2} y + 1 = 9

x^{2} + 11 x + 15 = 0

What are the solutions to the equation x(x-7)+12=0 ?
The solutions to the equation x(x-7)+12=0 are x=4,x=3
Find the zeros of x(x-7)+12=0
The zeros of x(x-7)+12=0 are x=4,x=3