What are the solutions to the equation (2x+1)(4x-3)=0 ?

The solutions to the equation (2x+1)(4x-3)=0 are x=-1/2 ,x= 3/4

Find the zeros of (2x+1)(4x-3)=0

The zeros of (2x+1)(4x-3)=0 are x=-1/2 ,x= 3/4

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(2 x + 1) (4 x - 3) = 0

x = - \frac{1}{2}, x = \frac{3}{4}

Decimal Notation

x = - 0.5, x = 0.75

Solution steps

(2 x + 1) (4 x - 3) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

2 x + 1 = 0 or 4 x - 3 = 0

Solve

2 x + 1 = 0 : x = - \frac{1}{2}

Solve

4 x - 3 = 0 : x = \frac{3}{4}

The solutions to the quadratic equation are:

x = - \frac{1}{2}, x = \frac{3}{4}

4 x - x^{2} + 8 = x^{2} - 2 x

- 2 x^{2} - 28 x + 30 = 0

(5 x^{2} + x)^{2} = (3 x + 3)^{2}

x, x^{2} + y^{2} - 6 x - 8 y = - 25

x - x^{2}

What are the solutions to the equation (2x+1)(4x-3)=0 ?
The solutions to the equation (2x+1)(4x-3)=0 are x=-1/2 ,x= 3/4
Find the zeros of (2x+1)(4x-3)=0
The zeros of (2x+1)(4x-3)=0 are x=-1/2 ,x= 3/4