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

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

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

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

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

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

Decimal Notation

x = - 1.5, x = - 0.75

Solution steps

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

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

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

Solve

2 x + 3 = 0 : x = - \frac{3}{2}

Solve

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

The solutions to the quadratic equation are:

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

\frac{3}{b - 8} + \frac{7}{b + 3}

(x^{2})^{\frac{3}{2}}

2^{k} \geq 125

\frac{( 2 x - 1 ) ( x + 4 )}{x - 4} > 0

2 > - 3 - 3 x \geq - 7

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