What are the solutions to the equation (x-7)(x+4)=-2 ?

The solutions to the equation (x-7)(x+4)=-2 are x=(3+sqrt(113))/2 ,x=(3-sqrt(113))/2

Find the zeros of (x-7)(x+4)=-2

The zeros of (x-7)(x+4)=-2 are x=(3+sqrt(113))/2 ,x=(3-sqrt(113))/2

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(x - 7) (x + 4) = - 2

x = \frac{3 + 113}{2}, x = \frac{3 - 113}{2}

Decimal Notation

x = 6.81507 \dots, x = - 3.81507 \dots

Solution steps

(x - 7) (x + 4) = - 2

Expand

(x - 7) (x + 4) : x^{2} - 3 x - 28

x^{2} - 3 x - 28 = - 2

Move

2

to the left side

x^{2} - 3 x - 26 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 \cdot 1 \cdot ( - 26 )}{2 \cdot 1}

(- 3)^{2} - 4 \cdot 1 \cdot (- 26) = 113

x_{1, 2} = \frac{- ( - 3 ) \pm 113}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 3 ) + 113}{2 \cdot 1}, x_{2} = \frac{- ( - 3 ) - 113}{2 \cdot 1}

x = \frac{- ( - 3 ) + 113}{2 \cdot 1} : \frac{3 + 113}{2}

x = \frac{- ( - 3 ) - 113}{2 \cdot 1} : \frac{3 - 113}{2}

The solutions to the quadratic equation are:

x = \frac{3 + 113}{2}, x = \frac{3 - 113}{2}

x, 3 x^{2} + 18 x - 3 y^{2} + 27 = 0

6 0^{2} + x^{2} = 6 1^{2}

2 x^{2} = 3 + 2 x

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

z, ln (x^{2} + y^{2}) + 3^{x y} + z^{2} + z = 0

What are the solutions to the equation (x-7)(x+4)=-2 ?
The solutions to the equation (x-7)(x+4)=-2 are x=(3+sqrt(113))/2 ,x=(3-sqrt(113))/2
Find the zeros of (x-7)(x+4)=-2
The zeros of (x-7)(x+4)=-2 are x=(3+sqrt(113))/2 ,x=(3-sqrt(113))/2