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

The solutions to the equation 2x^2-2x-3=0 are x=(1+sqrt(7))/2 ,x=(1-sqrt(7))/2

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

The zeros of 2x^2-2x-3=0 are x=(1+sqrt(7))/2 ,x=(1-sqrt(7))/2

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

x = \frac{1 + 7}{2}, x = \frac{1 - 7}{2}

Decimal Notation

x = 1.82287 \dots, x = - 0.82287 \dots

Solution steps

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

Solve with the quadratic formula

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

(- 2)^{2} - 4 \cdot 2 (- 3) = 27

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

Separate the solutions

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

x = \frac{- ( - 2 ) + 2 7}{2 \cdot 2} : \frac{1 + 7}{2}

x = \frac{- ( - 2 ) - 2 7}{2 \cdot 2} : \frac{1 - 7}{2}

The solutions to the quadratic equation are:

x = \frac{1 + 7}{2}, x = \frac{1 - 7}{2}

(- 5)^{1}

\frac{4 a ^{5}}{7 b ^{4}} \cdot \frac{2 b ^{2}}{2 a ^{4}}

m^{2} + 2 m - 35

\frac{9}{5} c + 32 = 68

\frac{- 2 + 5 i}{- 3 + 4 i}

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