What are the solutions to the equation (w+2)(w-3)=1 ?

The solutions to the equation (w+2)(w-3)=1 are w=(1+sqrt(29))/2 ,w=(1-sqrt(29))/2

Find the zeros of (w+2)(w-3)=1

The zeros of (w+2)(w-3)=1 are w=(1+sqrt(29))/2 ,w=(1-sqrt(29))/2

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(w + 2) (w - 3) = 1

w = \frac{1 + 29}{2}, w = \frac{1 - 29}{2}

Decimal Notation

w = 3.19258 \dots, w = - 2.19258 \dots

Solution steps

(w + 2) (w - 3) = 1

Expand

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

w^{2} - w - 6 = 1

Move

1

to the left side

w^{2} - w - 7 = 0

Solve with the quadratic formula

w_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 \cdot 1 \cdot ( - 7 )}{2 \cdot 1}

(- 1)^{2} - 4 \cdot 1 \cdot (- 7) = 29

w_{1, 2} = \frac{- ( - 1 ) \pm 29}{2 \cdot 1}

Separate the solutions

w_{1} = \frac{- ( - 1 ) + 29}{2 \cdot 1}, w_{2} = \frac{- ( - 1 ) - 29}{2 \cdot 1}

w = \frac{- ( - 1 ) + 29}{2 \cdot 1} : \frac{1 + 29}{2}

w = \frac{- ( - 1 ) - 29}{2 \cdot 1} : \frac{1 - 29}{2}

The solutions to the quadratic equation are:

w = \frac{1 + 29}{2}, w = \frac{1 - 29}{2}

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

2 x^{2} - 30 = 2

x (x - 8) = 9

x, - x^{2} + \frac{36 y ^{2}}{25} = 0

x^{2} - 4 x + 4 = 0

What are the solutions to the equation (w+2)(w-3)=1 ?
The solutions to the equation (w+2)(w-3)=1 are w=(1+sqrt(29))/2 ,w=(1-sqrt(29))/2
Find the zeros of (w+2)(w-3)=1
The zeros of (w+2)(w-3)=1 are w=(1+sqrt(29))/2 ,w=(1-sqrt(29))/2