What are the solutions to the equation 9t^2-4t+1=5t^2 ?

The solutions to the equation 9t^2-4t+1=5t^2 are t= 1/2

Find the zeros of 9t^2-4t+1=5t^2

The zeros of 9t^2-4t+1=5t^2 are t= 1/2

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9 t^{2} - 4 t + 1 = 5 t^{2}

t = \frac{1}{2}

Decimal Notation

t = 0.5

Solution steps

9 t^{2} - 4 t + 1 = 5 t^{2}

Move

5 t^{2}

to the left side

4 t^{2} - 4 t + 1 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 \cdot 4 \cdot 1}{2 \cdot 4}

(- 4)^{2} - 4 \cdot 4 \cdot 1 = 0

t_{1, 2} = \frac{- ( - 4 ) \pm 0}{2 \cdot 4}

t = \frac{- ( - 4 )}{2 \cdot 4}

\frac{- ( - 4 )}{2 \cdot 4} = \frac{1}{2}

t = \frac{1}{2}

The solution to the quadratic equation is:

t = \frac{1}{2}

\frac{( x + 1 ) ( x - 2 )}{( x + 3 )} < 0

650 = 2 (650 - 2 x) + 2 x

- 3 x^{2} + 5 x - 1 = 0

4 x^{2} + 8 x = 1

e^{5 x} \cdot e^{5 x}

What are the solutions to the equation 9t^2-4t+1=5t^2 ?
The solutions to the equation 9t^2-4t+1=5t^2 are t= 1/2
Find the zeros of 9t^2-4t+1=5t^2
The zeros of 9t^2-4t+1=5t^2 are t= 1/2