What are the solutions to the equation (x-3)(x+4)=30 ?

The solutions to the equation (x-3)(x+4)=30 are x=6,x=-7

Find the zeros of (x-3)(x+4)=30

The zeros of (x-3)(x+4)=30 are x=6,x=-7

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(x - 3) (x + 4) = 30

x = 6, x = - 7

Solution steps

(x - 3) (x + 4) = 30

Expand

(x - 3) (x + 4) : x^{2} + x - 12

x^{2} + x - 12 = 30

Move

30

to the left side

x^{2} + x - 42 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 1 \cdot ( - 42 )}{2 \cdot 1}

1^{2} - 4 \cdot 1 \cdot (- 42) = 13

x_{1, 2} = \frac{- 1 \pm 13}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 1 + 13}{2 \cdot 1}, x_{2} = \frac{- 1 - 13}{2 \cdot 1}

x = \frac{- 1 + 13}{2 \cdot 1} : 6

x = \frac{- 1 - 13}{2 \cdot 1} : - 7

The solutions to the quadratic equation are:

x = 6, x = - 7

\frac{d}{6} < 6

\frac{1}{2} x < 8

78 = \frac{1}{2} (n^{2} - n)

(1 0^{1})^{3}

x^{3} - 27 x < 0

What are the solutions to the equation (x-3)(x+4)=30 ?
The solutions to the equation (x-3)(x+4)=30 are x=6,x=-7
Find the zeros of (x-3)(x+4)=30
The zeros of (x-3)(x+4)=30 are x=6,x=-7