What are the solutions to the equation (4x+5)(x+1)=0 ?

The solutions to the equation (4x+5)(x+1)=0 are x=-5/4 ,x=-1

Find the zeros of (4x+5)(x+1)=0

The zeros of (4x+5)(x+1)=0 are x=-5/4 ,x=-1

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(4 x + 5) (x + 1) = 0

x = - \frac{5}{4}, x = - 1

Decimal Notation

x = - 1.25, x = - 1

Solution steps

(4 x + 5) (x + 1) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

4 x + 5 = 0 or x + 1 = 0

Solve

4 x + 5 = 0 : x = - \frac{5}{4}

Solve

x + 1 = 0 : x = - 1

The solutions to the quadratic equation are:

x = - \frac{5}{4}, x = - 1

ln (4) = x

4 m^{2} - 4 m + 1

lo g_{3} (5)

\frac{3 m ^{- 3} r ^{4} p ^{2}}{12 t ^{4}}

4 x^{2} - 6 x + 2

What are the solutions to the equation (4x+5)(x+1)=0 ?
The solutions to the equation (4x+5)(x+1)=0 are x=-5/4 ,x=-1
Find the zeros of (4x+5)(x+1)=0
The zeros of (4x+5)(x+1)=0 are x=-5/4 ,x=-1