What are the solutions to the equation (5x-8)(x+2)=0 ?

The solutions to the equation (5x-8)(x+2)=0 are x= 8/5 ,x=-2

Find the zeros of (5x-8)(x+2)=0

The zeros of (5x-8)(x+2)=0 are x= 8/5 ,x=-2

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(5 x - 8) (x + 2) = 0

x = \frac{8}{5}, x = - 2

Decimal Notation

x = 1.6, x = - 2

Solution steps

(5 x - 8) (x + 2) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

5 x - 8 = 0 or x + 2 = 0

Solve

5 x - 8 = 0 : x = \frac{8}{5}

Solve

x + 2 = 0 : x = - 2

The solutions to the quadratic equation are:

x = \frac{8}{5}, x = - 2

R, \frac{( π - θ ) R ^{2}}{2} = \frac{θ R ^{2}}{2}

\frac{1}{12500} + \frac{3 θ}{125} + θ^{2} = 0

x^{2} = \frac{x ^{2}}{8}

5 = (2 - h)^{2} + (4 - 2 h)^{2}

x, (x - \frac{a}{2})^{2} + y^{2} = \frac{a ^{2}}{4}

What are the solutions to the equation (5x-8)(x+2)=0 ?
The solutions to the equation (5x-8)(x+2)=0 are x= 8/5 ,x=-2
Find the zeros of (5x-8)(x+2)=0
The zeros of (5x-8)(x+2)=0 are x= 8/5 ,x=-2