What are the solutions to the equation x^2+4x-40=-8 ?

The solutions to the equation x^2+4x-40=-8 are x=4,x=-8

Find the zeros of x^2+4x-40=-8

The zeros of x^2+4x-40=-8 are x=4,x=-8

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x^{2} + 4 x - 40 = - 8

x = 4, x = - 8

Solution steps

x^{2} + 4 x - 40 = - 8

Move

8

to the left side

x^{2} + 4 x - 32 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 4 \pm 4 ^{2} - 4 \cdot 1 \cdot ( - 32 )}{2 \cdot 1}

4^{2} - 4 \cdot 1 \cdot (- 32) = 12

x_{1, 2} = \frac{- 4 \pm 12}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 4 + 12}{2 \cdot 1}, x_{2} = \frac{- 4 - 12}{2 \cdot 1}

x = \frac{- 4 + 12}{2 \cdot 1} : 4

x = \frac{- 4 - 12}{2 \cdot 1} : - 8

The solutions to the quadratic equation are:

x = 4, x = - 8

\frac{240 π}{180}

\frac{e ^{- 4} \cdot 4 ^{4}}{4 !}

5 x + 5 y

x + (x + 2) + (x + 4) = 24

∣ 3 x - 2 ∣ \geq 0

What are the solutions to the equation x^2+4x-40=-8 ?
The solutions to the equation x^2+4x-40=-8 are x=4,x=-8
Find the zeros of x^2+4x-40=-8
The zeros of x^2+4x-40=-8 are x=4,x=-8