What are the solutions to the equation (2x-3)(5x-1)=x^2+5 ?

The solutions to the equation (2x-3)(5x-1)=x^2+5 are x=2,x=-1/9

Find the zeros of (2x-3)(5x-1)=x^2+5

The zeros of (2x-3)(5x-1)=x^2+5 are x=2,x=-1/9

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(2 x - 3) (5 x - 1) = x^{2} + 5

x = 2, x = - \frac{1}{9}

Solution steps

(2 x - 3) (5 x - 1) = x^{2} + 5

Expand

(2 x - 3) (5 x - 1) : 10 x^{2} - 17 x + 3

10 x^{2} - 17 x + 3 = x^{2} + 5

Move

5

to the left side

10 x^{2} - 17 x - 2 = x^{2}

Move

x^{2}

to the left side

9 x^{2} - 17 x - 2 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 17 ) \pm ( - 17 ) ^{2} - 4 \cdot 9 ( - 2 )}{2 \cdot 9}

(- 17)^{2} - 4 \cdot 9 (- 2) = 19

x_{1, 2} = \frac{- ( - 17 ) \pm 19}{2 \cdot 9}

Separate the solutions

x_{1} = \frac{- ( - 17 ) + 19}{2 \cdot 9}, x_{2} = \frac{- ( - 17 ) - 19}{2 \cdot 9}

x = \frac{- ( - 17 ) + 19}{2 \cdot 9} : 2

x = \frac{- ( - 17 ) - 19}{2 \cdot 9} : - \frac{1}{9}

The solutions to the quadratic equation are:

x = 2, x = - \frac{1}{9}

w^{2} - 5 w - \frac{11}{4} = 0

4 x^{2} + 16 x + 4 = 0

(x + 6) (x + 3) = 4

3 p^{2} + 5 p = 68

n^{2} - 4 n + 3 = 0

What are the solutions to the equation (2x-3)(5x-1)=x^2+5 ?
The solutions to the equation (2x-3)(5x-1)=x^2+5 are x=2,x=-1/9
Find the zeros of (2x-3)(5x-1)=x^2+5
The zeros of (2x-3)(5x-1)=x^2+5 are x=2,x=-1/9