What are the solutions to the equation 2x^2+10x+26=8 ?

The solutions to the equation 2x^2+10x+26=8 are x=-5/2+i(sqrt(11))/2 ,x=-5/2-i(sqrt(11))/2

Find the zeros of 2x^2+10x+26=8

The zeros of 2x^2+10x+26=8 are x=-5/2+i(sqrt(11))/2 ,x=-5/2-i(sqrt(11))/2

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2 x^{2} + 10 x + 26 = 8

x = - \frac{5}{2} + i \frac{11}{2}, x = - \frac{5}{2} - i \frac{11}{2}

Solution steps

2 x^{2} + 10 x + 26 = 8

Move

8

to the left side

2 x^{2} + 10 x + 18 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 10 \pm 1 0 ^{2} - 4 \cdot 2 \cdot 18}{2 \cdot 2}

Simplify

1 0^{2} - 4 \cdot 2 \cdot 18 : 211 i

x_{1, 2} = \frac{- 10 \pm 2 11 i}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 10 + 2 11 i}{2 \cdot 2}, x_{2} = \frac{- 10 - 2 11 i}{2 \cdot 2}

x = \frac{- 10 + 2 11 i}{2 \cdot 2} : - \frac{5}{2} + i \frac{11}{2}

x = \frac{- 10 - 2 11 i}{2 \cdot 2} : - \frac{5}{2} - i \frac{11}{2}

The solutions to the quadratic equation are:

x = - \frac{5}{2} + i \frac{11}{2}, x = - \frac{5}{2} - i \frac{11}{2}

v, f = \frac{m v ^{2}}{r}

z^{2} + 6 z = 10

\frac{( x ( x - 8 ) )}{4} + 2 = 0

x - 1 (x + 11)^{2} = 1

7 y^{2} + 3 = 199

What are the solutions to the equation 2x^2+10x+26=8 ?
The solutions to the equation 2x^2+10x+26=8 are x=-5/2+i(sqrt(11))/2 ,x=-5/2-i(sqrt(11))/2
Find the zeros of 2x^2+10x+26=8
The zeros of 2x^2+10x+26=8 are x=-5/2+i(sqrt(11))/2 ,x=-5/2-i(sqrt(11))/2