What are the solutions to the equation 2t^2-4t-3=0 ?

The solutions to the equation 2t^2-4t-3=0 are t=(2+sqrt(10))/2 ,t=(2-sqrt(10))/2

Find the zeros of 2t^2-4t-3=0

The zeros of 2t^2-4t-3=0 are t=(2+sqrt(10))/2 ,t=(2-sqrt(10))/2

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2 t^{2} - 4 t - 3 = 0

t = \frac{2 + 10}{2}, t = \frac{2 - 10}{2}

Decimal Notation

t = 2.58113 \dots, t = - 0.58113 \dots

Solution steps

2 t^{2} - 4 t - 3 = 0

Solve with the quadratic formula

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

(- 4)^{2} - 4 \cdot 2 (- 3) = 210

t_{1, 2} = \frac{- ( - 4 ) \pm 2 10}{2 \cdot 2}

Separate the solutions

t_{1} = \frac{- ( - 4 ) + 2 10}{2 \cdot 2}, t_{2} = \frac{- ( - 4 ) - 2 10}{2 \cdot 2}

t = \frac{- ( - 4 ) + 2 10}{2 \cdot 2} : \frac{2 + 10}{2}

t = \frac{- ( - 4 ) - 2 10}{2 \cdot 2} : \frac{2 - 10}{2}

The solutions to the quadratic equation are:

t = \frac{2 + 10}{2}, t = \frac{2 - 10}{2}

ln (6 x)

4 y^{2} = 3 - y

\frac{1}{4} (a + 2) = \frac{1}{6} (5 - a)

2 x^{2} - 22 = 0

\frac{x}{3} = - 8

What are the solutions to the equation 2t^2-4t-3=0 ?
The solutions to the equation 2t^2-4t-3=0 are t=(2+sqrt(10))/2 ,t=(2-sqrt(10))/2
Find the zeros of 2t^2-4t-3=0
The zeros of 2t^2-4t-3=0 are t=(2+sqrt(10))/2 ,t=(2-sqrt(10))/2