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

The solutions to the equation 2x^2+7x-3=0 are x=(-7+sqrt(73))/4 ,x=(-7-sqrt(73))/4

Find the zeros of 2x^2+7x-3=0

The zeros of 2x^2+7x-3=0 are x=(-7+sqrt(73))/4 ,x=(-7-sqrt(73))/4

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2 x^{2} + 7 x - 3 = 0

x = \frac{- 7 + 73}{4}, x = \frac{- 7 - 73}{4}

Solution steps

2 x^{2} + 7 x - 3 = 0

Solve with the quadratic formula

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

7^{2} - 4 \cdot 2 (- 3) = 73

x_{1, 2} = \frac{- 7 \pm 73}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 7 + 73}{2 \cdot 2}, x_{2} = \frac{- 7 - 73}{2 \cdot 2}

x = \frac{- 7 + 73}{2 \cdot 2} : \frac{- 7 + 73}{4}

x = \frac{- 7 - 73}{2 \cdot 2} : \frac{- 7 - 73}{4}

The solutions to the quadratic equation are:

x = \frac{- 7 + 73}{4}, x = \frac{- 7 - 73}{4}

lo g_{6} (12) + lo g_{6} (3)

96 - 163

\frac{1}{5} y + 11 = 3 - \frac{3}{5} y

a^{2} b^{2} c^{2} - a^{2} c^{2} x^{2} + a^{2} c^{2} y^{2}

\frac{8 ( - 6 ) - 5}{- 5 ( - 6 ) + 4}

What are the solutions to the equation 2x^2+7x-3=0 ?
The solutions to the equation 2x^2+7x-3=0 are x=(-7+sqrt(73))/4 ,x=(-7-sqrt(73))/4
Find the zeros of 2x^2+7x-3=0
The zeros of 2x^2+7x-3=0 are x=(-7+sqrt(73))/4 ,x=(-7-sqrt(73))/4