What are the solutions to the equation x(2x-7)=4(x-3) ?

The solutions to the equation x(2x-7)=4(x-3) are x=4,x= 3/2

Find the zeros of x(2x-7)=4(x-3)

The zeros of x(2x-7)=4(x-3) are x=4,x= 3/2

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x (2 x - 7) = 4 (x - 3)

x = 4, x = \frac{3}{2}

Decimal Notation

x = 4, x = 1.5

Solution steps

x (2 x - 7) = 4 (x - 3)

Expand

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

Expand

4 (x - 3) : 4 x - 12

2 x^{2} - 7 x = 4 x - 12

Move

12

to the left side

2 x^{2} - 7 x + 12 = 4 x

Move

4 x

to the left side

2 x^{2} - 11 x + 12 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 11 ) \pm ( - 11 ) ^{2} - 4 \cdot 2 \cdot 12}{2 \cdot 2}

(- 11)^{2} - 4 \cdot 2 \cdot 12 = 5

x_{1, 2} = \frac{- ( - 11 ) \pm 5}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- ( - 11 ) + 5}{2 \cdot 2}, x_{2} = \frac{- ( - 11 ) - 5}{2 \cdot 2}

x = \frac{- ( - 11 ) + 5}{2 \cdot 2} : 4

x = \frac{- ( - 11 ) - 5}{2 \cdot 2} : \frac{3}{2}

The solutions to the quadratic equation are:

x = 4, x = \frac{3}{2}

x^{2} - 2 x - 240 = 0

x^{2} - 2 (k - 3) x + 4 k = 0

68.2 = 39.2 t - 4.9 t^{2}

(36 - R)^{2} + (18 - R)^{2} = R^{2}

5 x^{2} - 20 x - 12 = 0

What are the solutions to the equation x(2x-7)=4(x-3) ?
The solutions to the equation x(2x-7)=4(x-3) are x=4,x= 3/2
Find the zeros of x(2x-7)=4(x-3)
The zeros of x(2x-7)=4(x-3) are x=4,x= 3/2