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

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

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

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

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(4 c + 2) (2 c - 7) = 0

c = - \frac{1}{2}, c = \frac{7}{2}

Decimal Notation

c = - 0.5, c = 3.5

Solution steps

(4 c + 2) (2 c - 7) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

4 c + 2 = 0 or 2 c - 7 = 0

Solve

4 c + 2 = 0 : c = - \frac{1}{2}

Solve

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

The solutions to the quadratic equation are:

c = - \frac{1}{2}, c = \frac{7}{2}

3 - 6 x^{2} = 0

y^{2} - 10 y + 9 = 0

x^{2} + 11 x + 36 = 0

x^{2} - x = x + 15

2 x^{2} + 3 x - 9 = 0

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