What are the solutions to the equation (x-4)^2+5x(x-1)=16 ?

The solutions to the equation (x-4)^2+5x(x-1)=16 are x= 13/6 ,x=0

Find the zeros of (x-4)^2+5x(x-1)=16

The zeros of (x-4)^2+5x(x-1)=16 are x= 13/6 ,x=0

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(x - 4)^{2} + 5 x (x - 1) = 16

x = \frac{13}{6}, x = 0

Decimal Notation

x = 2.16666 \dots, x = 0

Solution steps

(x - 4)^{2} + 5 x (x - 1) = 16

Expand

(x - 4)^{2} + 5 x (x - 1) : 6 x^{2} - 13 x + 16

6 x^{2} - 13 x + 16 = 16

Move

16

to the left side

6 x^{2} - 13 x = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 13 ) \pm ( - 13 ) ^{2} - 4 \cdot 6 \cdot 0}{2 \cdot 6}

(- 13)^{2} - 4 \cdot 6 \cdot 0 = 13

x_{1, 2} = \frac{- ( - 13 ) \pm 13}{2 \cdot 6}

Separate the solutions

x_{1} = \frac{- ( - 13 ) + 13}{2 \cdot 6}, x_{2} = \frac{- ( - 13 ) - 13}{2 \cdot 6}

x = \frac{- ( - 13 ) + 13}{2 \cdot 6} : \frac{13}{6}

x = \frac{- ( - 13 ) - 13}{2 \cdot 6} : 0

The solutions to the quadratic equation are:

x = \frac{13}{6}, x = 0

1 5^{2} = 2 0^{2} + x^{2} - 2 \cdot 20 \cdot x \cdot 0.75

2 u^{2} + u - 1 = 0

s^{2} + 12 s + 400 = 0

x^{2} + 37 x = 0

- 0.5 a^{2} + 243 a - 3280.5 = 0

What are the solutions to the equation (x-4)^2+5x(x-1)=16 ?
The solutions to the equation (x-4)^2+5x(x-1)=16 are x= 13/6 ,x=0
Find the zeros of (x-4)^2+5x(x-1)=16
The zeros of (x-4)^2+5x(x-1)=16 are x= 13/6 ,x=0