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

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

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

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

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5 x (x - 1) = 2 (x + 3)

x = 2, x = - \frac{3}{5}

Decimal

x = 2, x = - 0.6

Solution steps

5 x (x - 1) = 2 (x + 3)

Expand

5 x (x - 1) : 5 x^{2} - 5 x

Expand

2 (x + 3) : 2 x + 6

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

Move

6

to the left side

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

Move

2 x

to the left side

5 x^{2} - 7 x - 6 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 7 ) \pm ( - 7 ) ^{2} - 4 \cdot 5 ( - 6 )}{2 \cdot 5}

(- 7)^{2} - 4 \cdot 5 (- 6) = 13

x_{1, 2} = \frac{- ( - 7 ) \pm 13}{2 \cdot 5}

Separate the solutions

x_{1} = \frac{- ( - 7 ) + 13}{2 \cdot 5}, x_{2} = \frac{- ( - 7 ) - 13}{2 \cdot 5}

x = \frac{- ( - 7 ) + 13}{2 \cdot 5} : 2

x = \frac{- ( - 7 ) - 13}{2 \cdot 5} : - \frac{3}{5}

The solutions to the quadratic equation are:

x = 2, x = - \frac{3}{5}

5 = - 16 t^{2} + 22 t + 6

x^{2} - 2 x = 124

2 x^{2} - 9 x + 35 = 0

x^{2} - 7 x = x - 13

4 x^{2} = 4 x + 1

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