What are the solutions to the equation (4x+1)(2x-7)=0 ?

The solutions to the equation (4x+1)(2x-7)=0 are x=-1/4 ,x= 7/2

Find the zeros of (4x+1)(2x-7)=0

The zeros of (4x+1)(2x-7)=0 are x=-1/4 ,x= 7/2

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(4 x + 1) (2 x - 7) = 0

x = - \frac{1}{4}, x = \frac{7}{2}

Decimal Notation

x = - 0.25, x = 3.5

Solution steps

(4 x + 1) (2 x - 7) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

4 x + 1 = 0 or 2 x - 7 = 0

Solve

4 x + 1 = 0 : x = - \frac{1}{4}

Solve

2 x - 7 = 0 : x = \frac{7}{2}

The solutions to the quadratic equation are:

x = - \frac{1}{4}, x = \frac{7}{2}

i^{- 203}

m^{2} + 14 m - 51

\frac{20 i}{3 + i}

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

3 + 4 x > 27

What are the solutions to the equation (4x+1)(2x-7)=0 ?
The solutions to the equation (4x+1)(2x-7)=0 are x=-1/4 ,x= 7/2
Find the zeros of (4x+1)(2x-7)=0
The zeros of (4x+1)(2x-7)=0 are x=-1/4 ,x= 7/2