What are the solutions to the equation (x+4)^2=3x+40 ?

The solutions to the equation (x+4)^2=3x+40 are x=3,x=-8

Find the zeros of (x+4)^2=3x+40

The zeros of (x+4)^2=3x+40 are x=3,x=-8

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(x + 4)^{2} = 3 x + 40

x = 3, x = - 8

Solution steps

(x + 4)^{2} = 3 x + 40

Expand

(x + 4)^{2} : x^{2} + 8 x + 16

x^{2} + 8 x + 16 = 3 x + 40

Move

40

to the left side

x^{2} + 8 x - 24 = 3 x

Move

3 x

to the left side

x^{2} + 5 x - 24 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 5 \pm 5 ^{2} - 4 \cdot 1 \cdot ( - 24 )}{2 \cdot 1}

5^{2} - 4 \cdot 1 \cdot (- 24) = 11

x_{1, 2} = \frac{- 5 \pm 11}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 5 + 11}{2 \cdot 1}, x_{2} = \frac{- 5 - 11}{2 \cdot 1}

x = \frac{- 5 + 11}{2 \cdot 1} : 3

x = \frac{- 5 - 11}{2 \cdot 1} : - 8

The solutions to the quadratic equation are:

x = 3, x = - 8

∣ x ∣ \leq 11

k - 8 = 11.8

x^{2} + 13 x - 14 > 0

3 x - 2 x = 2 x + 3 - 2 x

3 \cdot 1 0^{- 1}

What are the solutions to the equation (x+4)^2=3x+40 ?
The solutions to the equation (x+4)^2=3x+40 are x=3,x=-8
Find the zeros of (x+4)^2=3x+40
The zeros of (x+4)^2=3x+40 are x=3,x=-8