What are the solutions to the equation (10t+5)(t-2)=0 ?

The solutions to the equation (10t+5)(t-2)=0 are t=-1/2 ,t=2

Find the zeros of (10t+5)(t-2)=0

The zeros of (10t+5)(t-2)=0 are t=-1/2 ,t=2

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(10 t + 5) (t - 2) = 0

t = - \frac{1}{2}, t = 2

Decimal Notation

t = - 0.5, t = 2

Solution steps

(10 t + 5) (t - 2) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

10 t + 5 = 0 or t - 2 = 0

Solve

10 t + 5 = 0 : t = - \frac{1}{2}

Solve

t - 2 = 0 : t = 2

The solutions to the quadratic equation are:

t = - \frac{1}{2}, t = 2

\frac{( \frac{1}{3} )}{3}

- 6 n > 54

\frac{5}{6}

x^{2} - 4 x + 0 = 0

x^{2} = 13 x - 30

What are the solutions to the equation (10t+5)(t-2)=0 ?
The solutions to the equation (10t+5)(t-2)=0 are t=-1/2 ,t=2
Find the zeros of (10t+5)(t-2)=0
The zeros of (10t+5)(t-2)=0 are t=-1/2 ,t=2