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

The solutions to the equation 2t^2+7t+3=0 are t=-1/2 ,t=-3

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

The zeros of 2t^2+7t+3=0 are t=-1/2 ,t=-3

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

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

Decimal Notation

t = - 0.5, t = - 3

Solution steps

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

Solve with the quadratic formula

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

7^{2} - 4 \cdot 2 \cdot 3 = 5

t_{1, 2} = \frac{- 7 \pm 5}{2 \cdot 2}

Separate the solutions

t_{1} = \frac{- 7 + 5}{2 \cdot 2}, t_{2} = \frac{- 7 - 5}{2 \cdot 2}

t = \frac{- 7 + 5}{2 \cdot 2} : - \frac{1}{2}

t = \frac{- 7 - 5}{2 \cdot 2} : - 3

The solutions to the quadratic equation are:

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

(\frac{2}{3})^{2} + (x)^{2} = - (\frac{1}{4})^{2}

b^{2} - 4 b - 5

5 v^{2} - 26 v = 24

\frac{ln ( 2 )}{0.5}

∣ x + 2 ∣ \geq ∣ x - 3 ∣ - ∣ x - 2 ∣

What are the solutions to the equation 2t^2+7t+3=0 ?
The solutions to the equation 2t^2+7t+3=0 are t=-1/2 ,t=-3
Find the zeros of 2t^2+7t+3=0
The zeros of 2t^2+7t+3=0 are t=-1/2 ,t=-3