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

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

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

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

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

t = \frac{5}{2}, t = \frac{2}{5}

Decimal Notation

t = 2.5, t = 0.4

Solution steps

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

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

2 t - 5 = 0 or 5 t - 2 = 0

Solve

2 t - 5 = 0 : t = \frac{5}{2}

Solve

5 t - 2 = 0 : t = \frac{2}{5}

The solutions to the quadratic equation are:

t = \frac{5}{2}, t = \frac{2}{5}

- \frac{5}{8} x = 20

(- 11)^{2}

(3 - 2 i)^{3}

(7 x + 1)^{2} = (3 x + 3)^{2} + (8 x - 4)^{2}

21 \cdot 6

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