What are the solutions to the equation 2x^2-12x-182=0 ?

The solutions to the equation 2x^2-12x-182=0 are x=13,x=-7

Find the zeros of 2x^2-12x-182=0

The zeros of 2x^2-12x-182=0 are x=13,x=-7

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2 x^{2} - 12 x - 182 = 0

x = 13, x = - 7

Solution steps

2 x^{2} - 12 x - 182 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 12 ) \pm ( - 12 ) ^{2} - 4 \cdot 2 ( - 182 )}{2 \cdot 2}

(- 12)^{2} - 4 \cdot 2 (- 182) = 40

x_{1, 2} = \frac{- ( - 12 ) \pm 40}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- ( - 12 ) + 40}{2 \cdot 2}, x_{2} = \frac{- ( - 12 ) - 40}{2 \cdot 2}

x = \frac{- ( - 12 ) + 40}{2 \cdot 2} : 13

x = \frac{- ( - 12 ) - 40}{2 \cdot 2} : - 7

The solutions to the quadratic equation are:

x = 13, x = - 7

(3 x - 5)^{2} - (2 x + 1)^{2} = 24

x, f = a (x - h)^{2} + k

4 n^{2} - 9 = - n

x, (x - h)^{2} + (y - k)^{2} = r^{2}

4 w^{2} + 12 w - 44 = 0

What are the solutions to the equation 2x^2-12x-182=0 ?
The solutions to the equation 2x^2-12x-182=0 are x=13,x=-7
Find the zeros of 2x^2-12x-182=0
The zeros of 2x^2-12x-182=0 are x=13,x=-7