What are the solutions to the equation 2x^2+40x+32=0 ?

The solutions to the equation 2x^2+40x+32=0 are x=2(sqrt(21)-5),x=-2(5+sqrt(21))

Find the zeros of 2x^2+40x+32=0

The zeros of 2x^2+40x+32=0 are x=2(sqrt(21)-5),x=-2(5+sqrt(21))

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2 x^{2} + 40 x + 32 = 0

x = 2 (21 - 5), x = - 2 (5 + 21)

Decimal Notation

x = - 0.83484 \dots, x = - 19.16515 \dots

Solution steps

2 x^{2} + 40 x + 32 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 40 \pm 4 0 ^{2} - 4 \cdot 2 \cdot 32}{2 \cdot 2}

4 0^{2} - 4 \cdot 2 \cdot 32 = 821

x_{1, 2} = \frac{- 40 \pm 8 21}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 40 + 8 21}{2 \cdot 2}, x_{2} = \frac{- 40 - 8 21}{2 \cdot 2}

x = \frac{- 40 + 8 21}{2 \cdot 2} : 2 (21 - 5)

x = \frac{- 40 - 8 21}{2 \cdot 2} : - 2 (5 + 21)

The solutions to the quadratic equation are:

x = 2 (21 - 5), x = - 2 (5 + 21)

6 x^{2} - 4 x = - 2

r, V = π r^{2}

{\frac{3}{2} (p - 2)}^{2}

6^{2} + x^{2} = 100

x, x^{2} + y^{2} + z^{2} - 4 x = 0

What are the solutions to the equation 2x^2+40x+32=0 ?
The solutions to the equation 2x^2+40x+32=0 are x=2(sqrt(21)-5),x=-2(5+sqrt(21))
Find the zeros of 2x^2+40x+32=0
The zeros of 2x^2+40x+32=0 are x=2(sqrt(21)-5),x=-2(5+sqrt(21))