What are the solutions to the equation (x+1)(3x-5)=(5x-3)*4 ?

The solutions to the equation (x+1)(3x-5)=(5x-3)*4 are x=7,x= 1/3

Find the zeros of (x+1)(3x-5)=(5x-3)*4

The zeros of (x+1)(3x-5)=(5x-3)*4 are x=7,x= 1/3

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(x + 1) (3 x - 5) = (5 x - 3) \cdot 4

x = 7, x = \frac{1}{3}

Decimal Notation

x = 7, x = 0.33333 \dots

Solution steps

(x + 1) (3 x - 5) = (5 x - 3) \cdot 4

Expand

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

Expand

(5 x - 3) \cdot 4 : 20 x - 12

3 x^{2} - 2 x - 5 = 20 x - 12

Move

12

to the left side

3 x^{2} - 2 x + 7 = 20 x

Move

20 x

to the left side

3 x^{2} - 22 x + 7 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 22 ) \pm ( - 22 ) ^{2} - 4 \cdot 3 \cdot 7}{2 \cdot 3}

(- 22)^{2} - 4 \cdot 3 \cdot 7 = 20

x_{1, 2} = \frac{- ( - 22 ) \pm 20}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- ( - 22 ) + 20}{2 \cdot 3}, x_{2} = \frac{- ( - 22 ) - 20}{2 \cdot 3}

x = \frac{- ( - 22 ) + 20}{2 \cdot 3} : 7

x = \frac{- ( - 22 ) - 20}{2 \cdot 3} : \frac{1}{3}

The solutions to the quadratic equation are:

x = 7, x = \frac{1}{3}

r, z = r^{2}

0.25 [2.5 x + 1.5 (x - 4)] = - x \cdot x

2 y (3 - y) = 7

x^{2} - 4 x + 4 = - 25

x, w = 2 x^{2} + 3

What are the solutions to the equation (x+1)(3x-5)=(5x-3)*4 ?
The solutions to the equation (x+1)(3x-5)=(5x-3)*4 are x=7,x= 1/3
Find the zeros of (x+1)(3x-5)=(5x-3)*4
The zeros of (x+1)(3x-5)=(5x-3)*4 are x=7,x= 1/3