What are the solutions to the equation 13t^2-8t+1=-3t^2 ?

The solutions to the equation 13t^2-8t+1=-3t^2 are t= 1/4

Find the zeros of 13t^2-8t+1=-3t^2

The zeros of 13t^2-8t+1=-3t^2 are t= 1/4

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13 t^{2} - 8 t + 1 = - 3 t^{2}

t = \frac{1}{4}

Decimal Notation

t = 0.25

Solution steps

13 t^{2} - 8 t + 1 = - 3 t^{2}

Move

3 t^{2}

to the left side

16 t^{2} - 8 t + 1 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- ( - 8 ) \pm ( - 8 ) ^{2} - 4 \cdot 16 \cdot 1}{2 \cdot 16}

(- 8)^{2} - 4 \cdot 16 \cdot 1 = 0

t_{1, 2} = \frac{- ( - 8 ) \pm 0}{2 \cdot 16}

t = \frac{- ( - 8 )}{2 \cdot 16}

\frac{- ( - 8 )}{2 \cdot 16} = \frac{1}{4}

t = \frac{1}{4}

The solution to the quadratic equation is:

t = \frac{1}{4}

x, x^{2} + y^{2} + (z - 2)^{2} = 4

2 a x^{2} - (2 + a) b x + b^{2} = 0

x^{2} - 6 x = - 3 x + 28

x^{2} - 100 = 3 x^{2} + 96

s^{2} + 12 s + 936 = 0

What are the solutions to the equation 13t^2-8t+1=-3t^2 ?
The solutions to the equation 13t^2-8t+1=-3t^2 are t= 1/4
Find the zeros of 13t^2-8t+1=-3t^2
The zeros of 13t^2-8t+1=-3t^2 are t= 1/4