What are the solutions to the equation (x+5)(x-3)=9 ?

The solutions to the equation (x+5)(x-3)=9 are x=4,x=-6

Find the zeros of (x+5)(x-3)=9

The zeros of (x+5)(x-3)=9 are x=4,x=-6

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(x + 5) (x - 3) = 9

x = 4, x = - 6

Solution steps

(x + 5) (x - 3) = 9

Expand

(x + 5) (x - 3) : x^{2} + 2 x - 15

x^{2} + 2 x - 15 = 9

Move

9

to the left side

x^{2} + 2 x - 24 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 \cdot 1 \cdot ( - 24 )}{2 \cdot 1}

2^{2} - 4 \cdot 1 \cdot (- 24) = 10

x_{1, 2} = \frac{- 2 \pm 10}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 2 + 10}{2 \cdot 1}, x_{2} = \frac{- 2 - 10}{2 \cdot 1}

x = \frac{- 2 + 10}{2 \cdot 1} : 4

x = \frac{- 2 - 10}{2 \cdot 1} : - 6

The solutions to the quadratic equation are:

x = 4, x = - 6

2 x - 3 > 5 x + 4

∣ w - 4 ∣ > 4

320 = 144 t - 16 t^{2}

200 (\frac{( \frac{1 + 0.07}{12} ) ^{12 \cdot 35} - 1}{\frac{0.07}{12}})

x^{2} - 12 x - 1 = 0

What are the solutions to the equation (x+5)(x-3)=9 ?
The solutions to the equation (x+5)(x-3)=9 are x=4,x=-6
Find the zeros of (x+5)(x-3)=9
The zeros of (x+5)(x-3)=9 are x=4,x=-6