What are the solutions to the equation x^2-18(x-4)=0 ?

The solutions to the equation x^2-18(x-4)=0 are x=12,x=6

Find the zeros of x^2-18(x-4)=0

The zeros of x^2-18(x-4)=0 are x=12,x=6

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x^{2} - 18 (x - 4) = 0

x = 12, x = 6

Solution steps

x^{2} - 18 (x - 4) = 0

Expand

x^{2} - 18 (x - 4) : x^{2} - 18 x + 72

x^{2} - 18 x + 72 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 18 ) \pm ( - 18 ) ^{2} - 4 \cdot 1 \cdot 72}{2 \cdot 1}

(- 18)^{2} - 4 \cdot 1 \cdot 72 = 6

x_{1, 2} = \frac{- ( - 18 ) \pm 6}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 18 ) + 6}{2 \cdot 1}, x_{2} = \frac{- ( - 18 ) - 6}{2 \cdot 1}

x = \frac{- ( - 18 ) + 6}{2 \cdot 1} : 12

x = \frac{- ( - 18 ) - 6}{2 \cdot 1} : 6

The solutions to the quadratic equation are:

x = 12, x = 6

3 a^{2} - 8 a + 4 = 0

x^{2} + 2 x + 5 = - 5

x, f (x + 1) = 2 x^{2} + x - 1

63 (p^{2} - 1) = 32 p

- 5 x^{2} - 67 x - 100 = - 7 x

What are the solutions to the equation x^2-18(x-4)=0 ?
The solutions to the equation x^2-18(x-4)=0 are x=12,x=6
Find the zeros of x^2-18(x-4)=0
The zeros of x^2-18(x-4)=0 are x=12,x=6