What are the solutions to the equation (x+4)(x-5)+3=0 ?

The solutions to the equation (x+4)(x-5)+3=0 are x=(1+sqrt(69))/2 ,x=(1-sqrt(69))/2

Find the zeros of (x+4)(x-5)+3=0

The zeros of (x+4)(x-5)+3=0 are x=(1+sqrt(69))/2 ,x=(1-sqrt(69))/2

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(x + 4) (x - 5) + 3 = 0

x = \frac{1 + 69}{2}, x = \frac{1 - 69}{2}

Decimal Notation

x = 4.65331 \dots, x = - 3.65331 \dots

Solution steps

(x + 4) (x - 5) + 3 = 0

Expand

(x + 4) (x - 5) + 3 : x^{2} - x - 17

x^{2} - x - 17 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 \cdot 1 \cdot ( - 17 )}{2 \cdot 1}

(- 1)^{2} - 4 \cdot 1 \cdot (- 17) = 69

x_{1, 2} = \frac{- ( - 1 ) \pm 69}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 1 ) + 69}{2 \cdot 1}, x_{2} = \frac{- ( - 1 ) - 69}{2 \cdot 1}

x = \frac{- ( - 1 ) + 69}{2 \cdot 1} : \frac{1 + 69}{2}

x = \frac{- ( - 1 ) - 69}{2 \cdot 1} : \frac{1 - 69}{2}

The solutions to the quadratic equation are:

x = \frac{1 + 69}{2}, x = \frac{1 - 69}{2}

3 x^{2} - 12 + 9 = 0

x, z = \frac{1}{2} (x^{2} + y^{2})

4 (- 3 k + 25)^{2} (1 - k) = (- 6 k + 29)^{2} (- 5 - k)

x^{2} + 8 x - 945 = 0

- x^{2} + 4 x = 7

What are the solutions to the equation (x+4)(x-5)+3=0 ?
The solutions to the equation (x+4)(x-5)+3=0 are x=(1+sqrt(69))/2 ,x=(1-sqrt(69))/2
Find the zeros of (x+4)(x-5)+3=0
The zeros of (x+4)(x-5)+3=0 are x=(1+sqrt(69))/2 ,x=(1-sqrt(69))/2