What are the solutions to the equation (x+3)^2-x=20 ?

The solutions to the equation (x+3)^2-x=20 are x=(-5+sqrt(69))/2 ,x=(-5-sqrt(69))/2

Find the zeros of (x+3)^2-x=20

The zeros of (x+3)^2-x=20 are x=(-5+sqrt(69))/2 ,x=(-5-sqrt(69))/2

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(x + 3)^{2} - x = 20

x = \frac{- 5 + 69}{2}, x = \frac{- 5 - 69}{2}

Decimal

x = 1.65331 \dots, x = - 6.65331 \dots

Solution steps

(x + 3)^{2} - x = 20

Expand

(x + 3)^{2} - x : x^{2} + 5 x + 9

x^{2} + 5 x + 9 = 20

Move

20

to the left side

x^{2} + 5 x - 11 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 5 \pm 5 ^{2} - 4 \cdot 1 \cdot ( - 11 )}{2 \cdot 1}

5^{2} - 4 \cdot 1 \cdot (- 11) = 69

x_{1, 2} = \frac{- 5 \pm 69}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 5 + 69}{2 \cdot 1}, x_{2} = \frac{- 5 - 69}{2 \cdot 1}

x = \frac{- 5 + 69}{2 \cdot 1} : \frac{- 5 + 69}{2}

x = \frac{- 5 - 69}{2 \cdot 1} : \frac{- 5 - 69}{2}

The solutions to the quadratic equation are:

x = \frac{- 5 + 69}{2}, x = \frac{- 5 - 69}{2}

x = - 4 x^{2} - 4

x^{2} + 24 = 0

a^{2} = 121

17 w^{2} = 10625

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

What are the solutions to the equation (x+3)^2-x=20 ?
The solutions to the equation (x+3)^2-x=20 are x=(-5+sqrt(69))/2 ,x=(-5-sqrt(69))/2
Find the zeros of (x+3)^2-x=20
The zeros of (x+3)^2-x=20 are x=(-5+sqrt(69))/2 ,x=(-5-sqrt(69))/2