What are the solutions to the equation (x+1)(x-4)=6 ?

The solutions to the equation (x+1)(x-4)=6 are x=5,x=-2

Find the zeros of (x+1)(x-4)=6

The zeros of (x+1)(x-4)=6 are x=5,x=-2

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(x + 1) (x - 4) = 6

x = 5, x = - 2

Solution steps

(x + 1) (x - 4) = 6

Expand

(x + 1) (x - 4) : x^{2} - 3 x - 4

x^{2} - 3 x - 4 = 6

Move

6

to the left side

x^{2} - 3 x - 10 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 \cdot 1 \cdot ( - 10 )}{2 \cdot 1}

(- 3)^{2} - 4 \cdot 1 \cdot (- 10) = 7

x_{1, 2} = \frac{- ( - 3 ) \pm 7}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 3 ) + 7}{2 \cdot 1}, x_{2} = \frac{- ( - 3 ) - 7}{2 \cdot 1}

x = \frac{- ( - 3 ) + 7}{2 \cdot 1} : 5

x = \frac{- ( - 3 ) - 7}{2 \cdot 1} : - 2

The solutions to the quadratic equation are:

x = 5, x = - 2

x (2 x) = 1800

(3 x + 2)^{2} + (x + 4)^{2} = (2 x + 6)

5 x^{2} - 32.04 x - 120 = 0

- 6 x^{2} + 10 x + 10 = 0

- x^{2} + 4 x = 2 x - 7

What are the solutions to the equation (x+1)(x-4)=6 ?
The solutions to the equation (x+1)(x-4)=6 are x=5,x=-2
Find the zeros of (x+1)(x-4)=6
The zeros of (x+1)(x-4)=6 are x=5,x=-2