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

The solutions to the equation 7x(x-5)=3(x+1) are x=(19+sqrt(382))/7 ,x=(19-sqrt(382))/7

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

The zeros of 7x(x-5)=3(x+1) are x=(19+sqrt(382))/7 ,x=(19-sqrt(382))/7

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7 x (x - 5) = 3 (x + 1)

x = \frac{19 + 382}{7}, x = \frac{19 - 382}{7}

Decimal

x = 5.50640 \dots, x = - 0.07783 \dots

Solution steps

7 x (x - 5) = 3 (x + 1)

Expand

7 x (x - 5) : 7 x^{2} - 35 x

Expand

3 (x + 1) : 3 x + 3

7 x^{2} - 35 x = 3 x + 3

Move

3

to the left side

7 x^{2} - 35 x - 3 = 3 x

Move

3 x

to the left side

7 x^{2} - 38 x - 3 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 38 ) \pm ( - 38 ) ^{2} - 4 \cdot 7 ( - 3 )}{2 \cdot 7}

(- 38)^{2} - 4 \cdot 7 (- 3) = 2382

x_{1, 2} = \frac{- ( - 38 ) \pm 2 382}{2 \cdot 7}

Separate the solutions

x_{1} = \frac{- ( - 38 ) + 2 382}{2 \cdot 7}, x_{2} = \frac{- ( - 38 ) - 2 382}{2 \cdot 7}

x = \frac{- ( - 38 ) + 2 382}{2 \cdot 7} : \frac{19 + 382}{7}

x = \frac{- ( - 38 ) - 2 382}{2 \cdot 7} : \frac{19 - 382}{7}

The solutions to the quadratic equation are:

x = \frac{19 + 382}{7}, x = \frac{19 - 382}{7}

(6 x - 3)^{2} = 59

x^{2} - 6 x + 25

x^{2} + \frac{b x}{a} = - \frac{c}{a}

6 b^{2} + 6 b - 6 = b

x^{2} - 4 x - 12 = 0

What are the solutions to the equation 7x(x-5)=3(x+1) ?
The solutions to the equation 7x(x-5)=3(x+1) are x=(19+sqrt(382))/7 ,x=(19-sqrt(382))/7
Find the zeros of 7x(x-5)=3(x+1)
The zeros of 7x(x-5)=3(x+1) are x=(19+sqrt(382))/7 ,x=(19-sqrt(382))/7