What are the solutions to the equation (x-2)(x+5)=3 ?

The solutions to the equation (x-2)(x+5)=3 are x=(-3+sqrt(61))/2 ,x=(-3-sqrt(61))/2

Find the zeros of (x-2)(x+5)=3

The zeros of (x-2)(x+5)=3 are x=(-3+sqrt(61))/2 ,x=(-3-sqrt(61))/2

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(x - 2) (x + 5) = 3

x = \frac{- 3 + 61}{2}, x = \frac{- 3 - 61}{2}

Decimal Notation

x = 2.40512 \dots, x = - 5.40512 \dots

Solution steps

(x - 2) (x + 5) = 3

Expand

(x - 2) (x + 5) : x^{2} + 3 x - 10

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

Move

3

to the left side

x^{2} + 3 x - 13 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 3 \pm 3 ^{2} - 4 \cdot 1 \cdot ( - 13 )}{2 \cdot 1}

3^{2} - 4 \cdot 1 \cdot (- 13) = 61

x_{1, 2} = \frac{- 3 \pm 61}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 3 + 61}{2 \cdot 1}, x_{2} = \frac{- 3 - 61}{2 \cdot 1}

x = \frac{- 3 + 61}{2 \cdot 1} : \frac{- 3 + 61}{2}

x = \frac{- 3 - 61}{2 \cdot 1} : \frac{- 3 - 61}{2}

The solutions to the quadratic equation are:

x = \frac{- 3 + 61}{2}, x = \frac{- 3 - 61}{2}

x^{2} - 21 x - 72 = 0

81 x^{2} - 18 x + 1 = 0

6 (x^{2} - 1) = 3 (x^{2} + x) - 1

x^{2} = - 90

x^{2} + (x + 2)^{2} = 212

What are the solutions to the equation (x-2)(x+5)=3 ?
The solutions to the equation (x-2)(x+5)=3 are x=(-3+sqrt(61))/2 ,x=(-3-sqrt(61))/2
Find the zeros of (x-2)(x+5)=3
The zeros of (x-2)(x+5)=3 are x=(-3+sqrt(61))/2 ,x=(-3-sqrt(61))/2