What are the solutions to the equation 3x(2x-5)=-4x-3 ?

The solutions to the equation 3x(2x-5)=-4x-3 are x= 3/2 ,x= 1/3

Find the zeros of 3x(2x-5)=-4x-3

The zeros of 3x(2x-5)=-4x-3 are x= 3/2 ,x= 1/3

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3 x (2 x - 5) = - 4 x - 3

x = \frac{3}{2}, x = \frac{1}{3}

Decimal Notation

x = 1.5, x = 0.33333 \dots

Solution steps

3 x (2 x - 5) = - 4 x - 3

Expand

3 x (2 x - 5) : 6 x^{2} - 15 x

6 x^{2} - 15 x = - 4 x - 3

Move

3

to the left side

6 x^{2} - 15 x + 3 = - 4 x

Move

4 x

to the left side

6 x^{2} - 11 x + 3 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 11 ) \pm ( - 11 ) ^{2} - 4 \cdot 6 \cdot 3}{2 \cdot 6}

(- 11)^{2} - 4 \cdot 6 \cdot 3 = 7

x_{1, 2} = \frac{- ( - 11 ) \pm 7}{2 \cdot 6}

Separate the solutions

x_{1} = \frac{- ( - 11 ) + 7}{2 \cdot 6}, x_{2} = \frac{- ( - 11 ) - 7}{2 \cdot 6}

x = \frac{- ( - 11 ) + 7}{2 \cdot 6} : \frac{3}{2}

x = \frac{- ( - 11 ) - 7}{2 \cdot 6} : \frac{1}{3}

The solutions to the quadratic equation are:

x = \frac{3}{2}, x = \frac{1}{3}

12 x^{2} + 6 x - 18 = 0

20 x^{2} + 25 x + 5 = 0

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

c, a = b + c^{2} d

x, x^{2} + 4 y^{2} = 0

What are the solutions to the equation 3x(2x-5)=-4x-3 ?
The solutions to the equation 3x(2x-5)=-4x-3 are x= 3/2 ,x= 1/3
Find the zeros of 3x(2x-5)=-4x-3
The zeros of 3x(2x-5)=-4x-3 are x= 3/2 ,x= 1/3