What are the solutions to the equation 8x^2+40x+50=0 ?

The solutions to the equation 8x^2+40x+50=0 are x=-5/2

Find the zeros of 8x^2+40x+50=0

The zeros of 8x^2+40x+50=0 are x=-5/2

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8 x^{2} + 40 x + 50 = 0

x = - \frac{5}{2}

Decimal Notation

x = - 2.5

Solution steps

8 x^{2} + 40 x + 50 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 40 \pm 4 0 ^{2} - 4 \cdot 8 \cdot 50}{2 \cdot 8}

4 0^{2} - 4 \cdot 8 \cdot 50 = 0

x_{1, 2} = \frac{- 40 \pm 0}{2 \cdot 8}

x = \frac{- 40}{2 \cdot 8}

\frac{- 40}{2 \cdot 8} = - \frac{5}{2}

x = - \frac{5}{2}

The solution to the quadratic equation is:

x = - \frac{5}{2}

x^{2} + 4 x + \frac{7}{4} = 0

6 x^{2} + 3 = 11 x

(x - 4) (x + 4) + 6 (x + 4) = 3

8 p^{2} + 72 = 0

11 x^{2} = - 10 x - 11

What are the solutions to the equation 8x^2+40x+50=0 ?
The solutions to the equation 8x^2+40x+50=0 are x=-5/2
Find the zeros of 8x^2+40x+50=0
The zeros of 8x^2+40x+50=0 are x=-5/2