What are the solutions to the equation 4x^2+20x-56=0 ?

The solutions to the equation 4x^2+20x-56=0 are x=2,x=-7

Find the zeros of 4x^2+20x-56=0

The zeros of 4x^2+20x-56=0 are x=2,x=-7

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4 x^{2} + 20 x - 56 = 0

x = 2, x = - 7

Solution steps

4 x^{2} + 20 x - 56 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 20 \pm 2 0 ^{2} - 4 \cdot 4 ( - 56 )}{2 \cdot 4}

2 0^{2} - 4 \cdot 4 (- 56) = 36

x_{1, 2} = \frac{- 20 \pm 36}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- 20 + 36}{2 \cdot 4}, x_{2} = \frac{- 20 - 36}{2 \cdot 4}

x = \frac{- 20 + 36}{2 \cdot 4} : 2

x = \frac{- 20 - 36}{2 \cdot 4} : - 7

The solutions to the quadratic equation are:

x = 2, x = - 7

\frac{2 + 8}{2}

24 \geq - 3 b

\frac{3 - 2 ( x - 2 )}{6} - \frac{1 + 3 ( 2 - x )}{4} \leq - \frac{3 x - 1}{12}

∣ 4 x - 2 ∣ = ∣ x - 1 ∣

2 x^{2} + 3 + 4 x^{3} + 1

What are the solutions to the equation 4x^2+20x-56=0 ?
The solutions to the equation 4x^2+20x-56=0 are x=2,x=-7
Find the zeros of 4x^2+20x-56=0
The zeros of 4x^2+20x-56=0 are x=2,x=-7