What are the solutions to the equation 7(x-3)-5(x^2-1)=x^2-5(x+2) ?

The solutions to the equation 7(x-3)-5(x^2-1)=x^2-5(x+2) are x=1

Find the zeros of 7(x-3)-5(x^2-1)=x^2-5(x+2)

The zeros of 7(x-3)-5(x^2-1)=x^2-5(x+2) are x=1

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

x = 1

Solution steps

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

Expand

7 (x - 3) - 5 (x^{2} - 1) : - 5 x^{2} + 7 x - 16

Expand

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

- 5 x^{2} + 7 x - 16 = x^{2} - 5 x - 10

Move

10

to the left side

- 5 x^{2} + 7 x - 6 = x^{2} - 5 x

Move

5 x

to the left side

- 5 x^{2} + 12 x - 6 = x^{2}

Move

x^{2}

to the left side

- 6 x^{2} + 12 x - 6 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 12 \pm 1 2 ^{2} - 4 ( - 6 ) ( - 6 )}{2 ( - 6 )}

1 2^{2} - 4 (- 6) (- 6) = 0

x_{1, 2} = \frac{- 12 \pm 0}{2 ( - 6 )}

x = \frac{- 12}{2 ( - 6 )}

\frac{- 12}{2 ( - 6 )} = 1

x = 1

The solution to the quadratic equation is:

x = 1

\frac{2 3}{2}

\frac{2}{2} - 1

30 x^{2} - 62 x + 28 = 0

24 x - 9

\frac{5}{6} + \frac{1}{2}

What are the solutions to the equation 7(x-3)-5(x^2-1)=x^2-5(x+2) ?
The solutions to the equation 7(x-3)-5(x^2-1)=x^2-5(x+2) are x=1
Find the zeros of 7(x-3)-5(x^2-1)=x^2-5(x+2)
The zeros of 7(x-3)-5(x^2-1)=x^2-5(x+2) are x=1