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

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

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

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

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

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

Decimal Notation

x = 2.76556 \dots, x = - 1.26556 \dots

Solution steps

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

Solve with the quadratic formula

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

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

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

(z + 7)^{2} = - 81

\frac{4 ( 0.999 )}{( 0.999 ) - 1}

2 x - 3 > 5 or x - 3 < - 1

\frac{2 - 3}{- 2 - 5}

0.4 \cdot 0.9 = x \cdot 1.1

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