What are the solutions to the equation 40x^1+8x^2=800 ?

The solutions to the equation 40x^1+8x^2=800 are x=(5(sqrt(17)-1))/2 ,x=-(5(1+sqrt(17)))/2

Find the zeros of 40x^1+8x^2=800

The zeros of 40x^1+8x^2=800 are x=(5(sqrt(17)-1))/2 ,x=-(5(1+sqrt(17)))/2

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40 x^{1} + 8 x^{2} = 800

x = \frac{5 ( 17 - 1 )}{2}, x = - \frac{5 ( 1 + 17 )}{2}

Decimal Notation

x = 7.80776 \dots, x = - 12.80776 \dots

Solution steps

40 x^{1} + 8 x^{2} = 800

Refine

40 x + 8 x^{2} = 800

Move

800

to the left side

40 x + 8 x^{2} - 800 = 0

Write in the standard form

a x^{2} + b x + c = 0

8 x^{2} + 40 x - 800 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 40 \pm 4 0 ^{2} - 4 \cdot 8 ( - 800 )}{2 \cdot 8}

4 0^{2} - 4 \cdot 8 (- 800) = 4017

x_{1, 2} = \frac{- 40 \pm 40 17}{2 \cdot 8}

Separate the solutions

x_{1} = \frac{- 40 + 40 17}{2 \cdot 8}, x_{2} = \frac{- 40 - 40 17}{2 \cdot 8}

x = \frac{- 40 + 40 17}{2 \cdot 8} : \frac{5 ( 17 - 1 )}{2}

x = \frac{- 40 - 40 17}{2 \cdot 8} : - \frac{5 ( 1 + 17 )}{2}

The solutions to the quadratic equation are:

x = \frac{5 ( 17 - 1 )}{2}, x = - \frac{5 ( 1 + 17 )}{2}

8 1^{\frac{1}{2}}

(x - 6) (x + 0.7) = 0

∣ 2 x - 2 ∣ > 8

\frac{n ^{2} + 7 n}{4 n ^{2} + 28 n}

\frac{1}{6} < \frac{2 x - 13}{12} < \frac{2}{3}

What are the solutions to the equation 40x^1+8x^2=800 ?
The solutions to the equation 40x^1+8x^2=800 are x=(5(sqrt(17)-1))/2 ,x=-(5(1+sqrt(17)))/2
Find the zeros of 40x^1+8x^2=800
The zeros of 40x^1+8x^2=800 are x=(5(sqrt(17)-1))/2 ,x=-(5(1+sqrt(17)))/2