What are the solutions to the equation 8x^2+16x=40 ?

The solutions to the equation 8x^2+16x=40 are x=-1+sqrt(6),x=-1-sqrt(6)

Find the zeros of 8x^2+16x=40

The zeros of 8x^2+16x=40 are x=-1+sqrt(6),x=-1-sqrt(6)

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8 x^{2} + 16 x = 40

x = - 1 + 6, x = - 1 - 6

Decimal Notation

x = 1.44948 \dots, x = - 3.44948 \dots

Solution steps

8 x^{2} + 16 x = 40

Move

40

to the left side

8 x^{2} + 16 x - 40 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 16 \pm 1 6 ^{2} - 4 \cdot 8 ( - 40 )}{2 \cdot 8}

1 6^{2} - 4 \cdot 8 (- 40) = 166

x_{1, 2} = \frac{- 16 \pm 16 6}{2 \cdot 8}

Separate the solutions

x_{1} = \frac{- 16 + 16 6}{2 \cdot 8}, x_{2} = \frac{- 16 - 16 6}{2 \cdot 8}

x = \frac{- 16 + 16 6}{2 \cdot 8} : - 1 + 6

x = \frac{- 16 - 16 6}{2 \cdot 8} : - 1 - 6

The solutions to the quadratic equation are:

x = - 1 + 6, x = - 1 - 6

11 - 0.3 x = 0.83

- 7 + 5 (3 x - 4) = - 42

\frac{( x + 3 )}{2 x - 1} \geq 1

- \frac{1}{2} (- 4)

\frac{15 a ^{6} b c ^{4}}{35 a ^{2} c ^{4}}

What are the solutions to the equation 8x^2+16x=40 ?
The solutions to the equation 8x^2+16x=40 are x=-1+sqrt(6),x=-1-sqrt(6)
Find the zeros of 8x^2+16x=40
The zeros of 8x^2+16x=40 are x=-1+sqrt(6),x=-1-sqrt(6)