What are the solutions to the equation 2t^2-t-3=0 ?

The solutions to the equation 2t^2-t-3=0 are t= 3/2 ,t=-1

Find the zeros of 2t^2-t-3=0

The zeros of 2t^2-t-3=0 are t= 3/2 ,t=-1

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2 t^{2} - t - 3 = 0

t = \frac{3}{2}, t = - 1

Decimal Notation

t = 1.5, t = - 1

Solution steps

2 t^{2} - t - 3 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 \cdot 2 ( - 3 )}{2 \cdot 2}

(- 1)^{2} - 4 \cdot 2 (- 3) = 5

t_{1, 2} = \frac{- ( - 1 ) \pm 5}{2 \cdot 2}

Separate the solutions

t_{1} = \frac{- ( - 1 ) + 5}{2 \cdot 2}, t_{2} = \frac{- ( - 1 ) - 5}{2 \cdot 2}

t = \frac{- ( - 1 ) + 5}{2 \cdot 2} : \frac{3}{2}

t = \frac{- ( - 1 ) - 5}{2 \cdot 2} : - 1

The solutions to the quadratic equation are:

t = \frac{3}{2}, t = - 1

5 x^{2} + 10 = 10

4 x^{2} + 5 x - 7 = 0

x, \int y d y = x^{2}

x, (f \circ g) = x^{2}

4 y^{2} - 28 y + 49 = 0

What are the solutions to the equation 2t^2-t-3=0 ?
The solutions to the equation 2t^2-t-3=0 are t= 3/2 ,t=-1
Find the zeros of 2t^2-t-3=0
The zeros of 2t^2-t-3=0 are t= 3/2 ,t=-1