What are the solutions to the equation solvefor x,(x-3)(x+4)=18 ?

The solutions to the equation solvefor x,(x-3)(x+4)=18 are x=5,x=-6

Find the zeros of solvefor x,(x-3)(x+4)=18

The zeros of solvefor x,(x-3)(x+4)=18 are x=5,x=-6

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x, (x - 3) (x + 4) = 18

x = 5, x = - 6

Solution steps

(x - 3) (x + 4) = 18

Expand

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

x^{2} + x - 12 = 18

Move

18

to the left side

x^{2} + x - 30 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 1 \cdot ( - 30 )}{2 \cdot 1}

1^{2} - 4 \cdot 1 \cdot (- 30) = 11

x_{1, 2} = \frac{- 1 \pm 11}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 1 + 11}{2 \cdot 1}, x_{2} = \frac{- 1 - 11}{2 \cdot 1}

x = \frac{- 1 + 11}{2 \cdot 1} : 5

x = \frac{- 1 - 11}{2 \cdot 1} : - 6

The solutions to the quadratic equation are:

x = 5, x = - 6

(5 - x)^{2} + 3 = 0

x^{2} + 30 x - 2 = 0

16 x^{2} = - 9

x^{2} - 8 x - 84 = 0

x + 7 = x^{2} + 2 x + 1

What are the solutions to the equation solvefor x,(x-3)(x+4)=18 ?
The solutions to the equation solvefor x,(x-3)(x+4)=18 are x=5,x=-6
Find the zeros of solvefor x,(x-3)(x+4)=18
The zeros of solvefor x,(x-3)(x+4)=18 are x=5,x=-6