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}

Decimal Notation

x = 3.88600 \dots, x = - 0.38600 \dots

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}

\frac{2 - 6}{2 + 6}

z + (- z) = 4

∣ 2 (x - 1) + 12 ∣ \leq 10

5 x^{2} + 1 = 0

5 x < - 10

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