What are the solutions to the equation (x+5)(x-6)=-26 ?

The solutions to the equation (x+5)(x-6)=-26 are x=(1+sqrt(17))/2 ,x=(1-sqrt(17))/2

Find the zeros of (x+5)(x-6)=-26

The zeros of (x+5)(x-6)=-26 are x=(1+sqrt(17))/2 ,x=(1-sqrt(17))/2

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(x + 5) (x - 6) = - 26

x = \frac{1 + 17}{2}, x = \frac{1 - 17}{2}

Decimal Notation

x = 2.56155 \dots, x = - 1.56155 \dots

Solution steps

(x + 5) (x - 6) = - 26

Expand

(x + 5) (x - 6) : x^{2} - x - 30

x^{2} - x - 30 = - 26

Move

26

to the left side

x^{2} - x - 4 = 0

Solve with the quadratic formula

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

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

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

Separate the solutions

x_{1} = \frac{- ( - 1 ) + 17}{2 \cdot 1}, x_{2} = \frac{- ( - 1 ) - 17}{2 \cdot 1}

x = \frac{- ( - 1 ) + 17}{2 \cdot 1} : \frac{1 + 17}{2}

x = \frac{- ( - 1 ) - 17}{2 \cdot 1} : \frac{1 - 17}{2}

The solutions to the quadratic equation are:

x = \frac{1 + 17}{2}, x = \frac{1 - 17}{2}

x, 5 x^{2} - 6 x y + 5 y^{2} = 4

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

x = 9 - x^{2}

k^{2} - 11 k + 30 = 0

3 n^{2} - 2 n - 2 = 0

What are the solutions to the equation (x+5)(x-6)=-26 ?
The solutions to the equation (x+5)(x-6)=-26 are x=(1+sqrt(17))/2 ,x=(1-sqrt(17))/2
Find the zeros of (x+5)(x-6)=-26
The zeros of (x+5)(x-6)=-26 are x=(1+sqrt(17))/2 ,x=(1-sqrt(17))/2