What are the solutions to the equation 2t^2-3t-7=0 ?

The solutions to the equation 2t^2-3t-7=0 are t=(3+sqrt(65))/4 ,t=(3-sqrt(65))/4

Find the zeros of 2t^2-3t-7=0

The zeros of 2t^2-3t-7=0 are t=(3+sqrt(65))/4 ,t=(3-sqrt(65))/4

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2 t^{2} - 3 t - 7 = 0

t = \frac{3 + 65}{4}, t = \frac{3 - 65}{4}

Decimal Notation

t = 2.76556 \dots, t = - 1.26556 \dots

Solution steps

2 t^{2} - 3 t - 7 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 \cdot 2 ( - 7 )}{2 \cdot 2}

(- 3)^{2} - 4 \cdot 2 (- 7) = 65

t_{1, 2} = \frac{- ( - 3 ) \pm 65}{2 \cdot 2}

Separate the solutions

t_{1} = \frac{- ( - 3 ) + 65}{2 \cdot 2}, t_{2} = \frac{- ( - 3 ) - 65}{2 \cdot 2}

t = \frac{- ( - 3 ) + 65}{2 \cdot 2} : \frac{3 + 65}{4}

t = \frac{- ( - 3 ) - 65}{2 \cdot 2} : \frac{3 - 65}{4}

The solutions to the quadratic equation are:

t = \frac{3 + 65}{4}, t = \frac{3 - 65}{4}

x^{2} + 9 x = - 8

3 x^{1} + 2 x^{2} = 54

- 4 + 64 x^{2} = - 3

6 x^{2} - x - 12 = 0

201 = 0.1 w^{2} + 2.155 w + 19.985

What are the solutions to the equation 2t^2-3t-7=0 ?
The solutions to the equation 2t^2-3t-7=0 are t=(3+sqrt(65))/4 ,t=(3-sqrt(65))/4
Find the zeros of 2t^2-3t-7=0
The zeros of 2t^2-3t-7=0 are t=(3+sqrt(65))/4 ,t=(3-sqrt(65))/4