What are the solutions to the equation 2x^2-7x-32=0 ?

The solutions to the equation 2x^2-7x-32=0 are x=(7+sqrt(305))/4 ,x=(7-sqrt(305))/4

Find the zeros of 2x^2-7x-32=0

The zeros of 2x^2-7x-32=0 are x=(7+sqrt(305))/4 ,x=(7-sqrt(305))/4

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2 x^{2} - 7 x - 32 = 0

x = \frac{7 + 305}{4}, x = \frac{7 - 305}{4}

Decimal Notation

x = 6.11606 \dots, x = - 2.61606 \dots

Solution steps

2 x^{2} - 7 x - 32 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 7 ) \pm ( - 7 ) ^{2} - 4 \cdot 2 ( - 32 )}{2 \cdot 2}

(- 7)^{2} - 4 \cdot 2 (- 32) = 305

x_{1, 2} = \frac{- ( - 7 ) \pm 305}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- ( - 7 ) + 305}{2 \cdot 2}, x_{2} = \frac{- ( - 7 ) - 305}{2 \cdot 2}

x = \frac{- ( - 7 ) + 305}{2 \cdot 2} : \frac{7 + 305}{4}

x = \frac{- ( - 7 ) - 305}{2 \cdot 2} : \frac{7 - 305}{4}

The solutions to the quadratic equation are:

x = \frac{7 + 305}{4}, x = \frac{7 - 305}{4}

2 x^{2} + 5 x - 14 = 0

x^{2} - 5 x + 2 = 3 - 2 x

r^{2} - \frac{4}{9} = 0

- 4 v^{2} + 8 v - 4 = 0

2 (2 u^{2} - 4 u) + 3 = 0

What are the solutions to the equation 2x^2-7x-32=0 ?
The solutions to the equation 2x^2-7x-32=0 are x=(7+sqrt(305))/4 ,x=(7-sqrt(305))/4
Find the zeros of 2x^2-7x-32=0
The zeros of 2x^2-7x-32=0 are x=(7+sqrt(305))/4 ,x=(7-sqrt(305))/4