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

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

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

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

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

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

Decimal Notation

x = - 0.5, x = 3

Solution steps

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

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

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

Solve

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

Solve

x - 3 = 0 : x = 3

The solutions to the quadratic equation are:

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

(5 x + 1)^{2} + 3 = 10

2 (x + 3) (x - 2)^{3} + (x + 3)^{2} \cdot 3 (x - 2)^{2}

\frac{8 ( 3 ) - 3 ( 4 ) ^{2}}{3 ( 3 ) ^{2} - 8 ( 4 )}

722

7 a^{2} - 11 a + 4

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