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

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

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

The zeros of quadraticformula (x+2)(x-3)=0 are x=3,x=-2

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

x = 3, x = - 2

Solution steps

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

Expand

(x + 2) (x - 3) : x^{2} - x - 6

x^{2} - x - 6 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 \cdot 1 \cdot ( - 6 )}{2 \cdot 1}

(- 1)^{2} - 4 \cdot 1 \cdot (- 6) = 5

x_{1, 2} = \frac{- ( - 1 ) \pm 5}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 1 ) + 5}{2 \cdot 1}, x_{2} = \frac{- ( - 1 ) - 5}{2 \cdot 1}

x = \frac{- ( - 1 ) + 5}{2 \cdot 1} : 3

x = \frac{- ( - 1 ) - 5}{2 \cdot 1} : - 2

The solutions to the quadratic equation are:

x = 3, x = - 2

16 x^{3}

16 x^{2} = 56 x

z, 6 z + z - 4 z + 4 z - 5 z = 14

1 + tan (x)

\frac{( \frac{1}{x} - \frac{1}{3} )}{( x - 3 )}

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