What are the solutions to the equation quadraticformula (x-6)(x+2)=0 ?

The solutions to the equation quadraticformula (x-6)(x+2)=0 are x=6,x=-2

Find the zeros of quadraticformula (x-6)(x+2)=0

The zeros of quadraticformula (x-6)(x+2)=0 are x=6,x=-2

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(x - 6) (x + 2) = 0

x = 6, x = - 2

Solution steps

(x - 6) (x + 2) = 0

Expand

(x - 6) (x + 2) : x^{2} - 4 x - 12

x^{2} - 4 x - 12 = 0

Solve with the quadratic formula

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

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

x_{1, 2} = \frac{- ( - 4 ) \pm 8}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 4 ) + 8}{2 \cdot 1}, x_{2} = \frac{- ( - 4 ) - 8}{2 \cdot 1}

x = \frac{- ( - 4 ) + 8}{2 \cdot 1} : 6

x = \frac{- ( - 4 ) - 8}{2 \cdot 1} : - 2

The solutions to the quadratic equation are:

x = 6, x = - 2

\frac{36}{240}

x^{2} - 12 x + 27 = 0

40 x^{2}

\frac{( 2 x ^{4} y ^{- 4} z ^{- 3} )}{( 3 x ^{2} y ^{- 3} z ^{4} )}

x, \frac{7}{12} x = \frac{3}{14}

What are the solutions to the equation quadraticformula (x-6)(x+2)=0 ?
The solutions to the equation quadraticformula (x-6)(x+2)=0 are x=6,x=-2
Find the zeros of quadraticformula (x-6)(x+2)=0
The zeros of quadraticformula (x-6)(x+2)=0 are x=6,x=-2