What are the solutions to the equation 2w^2-5w-52=0 ?

The solutions to the equation 2w^2-5w-52=0 are w= 13/2 ,w=-4

Find the zeros of 2w^2-5w-52=0

The zeros of 2w^2-5w-52=0 are w= 13/2 ,w=-4

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2 w^{2} - 5 w - 52 = 0

w = \frac{13}{2}, w = - 4

Solution steps

2 w^{2} - 5 w - 52 = 0

Solve with the quadratic formula

w_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 \cdot 2 ( - 52 )}{2 \cdot 2}

(- 5)^{2} - 4 \cdot 2 (- 52) = 21

w_{1, 2} = \frac{- ( - 5 ) \pm 21}{2 \cdot 2}

Separate the solutions

w_{1} = \frac{- ( - 5 ) + 21}{2 \cdot 2}, w_{2} = \frac{- ( - 5 ) - 21}{2 \cdot 2}

w = \frac{- ( - 5 ) + 21}{2 \cdot 2} : \frac{13}{2}

w = \frac{- ( - 5 ) - 21}{2 \cdot 2} : - 4

The solutions to the quadratic equation are:

w = \frac{13}{2}, w = - 4

2^{\frac{168}{50}}

0 = - 5 t^{2} + 18 t + 16

(2 x + 1) + (3 x + 3) = 180

5 x^{2} - 5 x - 6 = 0

\frac{1}{3 ^{- 1}}

What are the solutions to the equation 2w^2-5w-52=0 ?
The solutions to the equation 2w^2-5w-52=0 are w= 13/2 ,w=-4
Find the zeros of 2w^2-5w-52=0
The zeros of 2w^2-5w-52=0 are w= 13/2 ,w=-4