What are the solutions to the equation 4x^2+140x-296=0 ?

The solutions to the equation 4x^2+140x-296=0 are x=2,x=-37

Find the zeros of 4x^2+140x-296=0

The zeros of 4x^2+140x-296=0 are x=2,x=-37

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4 x^{2} + 140 x - 296 = 0

x = 2, x = - 37

Solution steps

4 x^{2} + 140 x - 296 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 140 \pm 14 0 ^{2} - 4 \cdot 4 ( - 296 )}{2 \cdot 4}

14 0^{2} - 4 \cdot 4 (- 296) = 156

x_{1, 2} = \frac{- 140 \pm 156}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- 140 + 156}{2 \cdot 4}, x_{2} = \frac{- 140 - 156}{2 \cdot 4}

x = \frac{- 140 + 156}{2 \cdot 4} : 2

x = \frac{- 140 - 156}{2 \cdot 4} : - 37

The solutions to the quadratic equation are:

x = 2, x = - 37

x, x^{2} - y^{2} = a^{2}

- p^{2} + 5 p + 6 = - 2 p^{2}

(x - 64) (x + 64) = 0

x, 2 x^{2} + y^{2} = 4 C x

- 12 x^{2} + 11 x - 2 = 0

What are the solutions to the equation 4x^2+140x-296=0 ?
The solutions to the equation 4x^2+140x-296=0 are x=2,x=-37
Find the zeros of 4x^2+140x-296=0
The zeros of 4x^2+140x-296=0 are x=2,x=-37