What are the solutions to the equation (x-5)^2+8(x-5)+15=0 ?

The solutions to the equation (x-5)^2+8(x-5)+15=0 are x=2,x=0

Find the zeros of (x-5)^2+8(x-5)+15=0

The zeros of (x-5)^2+8(x-5)+15=0 are x=2,x=0

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(x - 5)^{2} + 8 (x - 5) + 15 = 0

x = 2, x = 0

Solution steps

(x - 5)^{2} + 8 (x - 5) + 15 = 0

Expand

(x - 5)^{2} + 8 (x - 5) + 15 : x^{2} - 2 x

x^{2} - 2 x = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 \cdot 1 \cdot 0}{2 \cdot 1}

(- 2)^{2} - 4 \cdot 1 \cdot 0 = 2

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

Separate the solutions

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

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

x = \frac{- ( - 2 ) - 2}{2 \cdot 1} : 0

The solutions to the quadratic equation are:

x = 2, x = 0

10 - x^{2} = - 27

x^{2} - 20 x + 51 = 0

5 x^{2} - 37 x - 24 = 0

a, b^{2} x^{2} + a^{2} y^{2} = a^{2} b^{2}

a, a (b - y) - a (b - 1) = a (a y - b)

What are the solutions to the equation (x-5)^2+8(x-5)+15=0 ?
The solutions to the equation (x-5)^2+8(x-5)+15=0 are x=2,x=0
Find the zeros of (x-5)^2+8(x-5)+15=0
The zeros of (x-5)^2+8(x-5)+15=0 are x=2,x=0