What are the solutions to the equation 2x^2-8x-42=0 ?

The solutions to the equation 2x^2-8x-42=0 are x=7,x=-3

Find the zeros of 2x^2-8x-42=0

The zeros of 2x^2-8x-42=0 are x=7,x=-3

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2 x^{2} - 8 x - 42 = 0

x = 7, x = - 3

Solution steps

2 x^{2} - 8 x - 42 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 8 ) \pm ( - 8 ) ^{2} - 4 \cdot 2 ( - 42 )}{2 \cdot 2}

(- 8)^{2} - 4 \cdot 2 (- 42) = 20

x_{1, 2} = \frac{- ( - 8 ) \pm 20}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- ( - 8 ) + 20}{2 \cdot 2}, x_{2} = \frac{- ( - 8 ) - 20}{2 \cdot 2}

x = \frac{- ( - 8 ) + 20}{2 \cdot 2} : 7

x = \frac{- ( - 8 ) - 20}{2 \cdot 2} : - 3

The solutions to the quadratic equation are:

x = 7, x = - 3

6 + 2 x^{2} - 3 x = 8 x^{2}

\frac{- 24 a ^{8}}{- 6 a ^{5}}

49 x^{2} + 84 x + 36

- 2 ∣ 3 x + 5 ∣ - 10 = - 18

b = \frac{13}{908} - \frac{5}{25} \cdot 908

What are the solutions to the equation 2x^2-8x-42=0 ?
The solutions to the equation 2x^2-8x-42=0 are x=7,x=-3
Find the zeros of 2x^2-8x-42=0
The zeros of 2x^2-8x-42=0 are x=7,x=-3