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

The solutions to the equation (5x+2)(x-1)=7x are x=(5+sqrt(35))/5 ,x=(5-sqrt(35))/5

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

The zeros of (5x+2)(x-1)=7x are x=(5+sqrt(35))/5 ,x=(5-sqrt(35))/5

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(5 x + 2) (x - 1) = 7 x

x = \frac{5 + 35}{5}, x = \frac{5 - 35}{5}

Decimal Notation

x = 2.18321 \dots, x = - 0.18321 \dots

Solution steps

(5 x + 2) (x - 1) = 7 x

Expand

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

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

Move

7 x

to the left side

5 x^{2} - 10 x - 2 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 10 ) \pm ( - 10 ) ^{2} - 4 \cdot 5 ( - 2 )}{2 \cdot 5}

(- 10)^{2} - 4 \cdot 5 (- 2) = 235

x_{1, 2} = \frac{- ( - 10 ) \pm 2 35}{2 \cdot 5}

Separate the solutions

x_{1} = \frac{- ( - 10 ) + 2 35}{2 \cdot 5}, x_{2} = \frac{- ( - 10 ) - 2 35}{2 \cdot 5}

x = \frac{- ( - 10 ) + 2 35}{2 \cdot 5} : \frac{5 + 35}{5}

x = \frac{- ( - 10 ) - 2 35}{2 \cdot 5} : \frac{5 - 35}{5}

The solutions to the quadratic equation are:

x = \frac{5 + 35}{5}, x = \frac{5 - 35}{5}

t^{2} + 3 t + 2 = 0

(x - 1)^{2} = 14

\frac{1}{4} (2 x - 12)^{2} + 8 = 4

5 x - x^{2}

0.32 x^{2} - 1.2 x + 1 = 0

What are the solutions to the equation (5x+2)(x-1)=7x ?
The solutions to the equation (5x+2)(x-1)=7x are x=(5+sqrt(35))/5 ,x=(5-sqrt(35))/5
Find the zeros of (5x+2)(x-1)=7x
The zeros of (5x+2)(x-1)=7x are x=(5+sqrt(35))/5 ,x=(5-sqrt(35))/5