What are the solutions to the equation x^2+2x=10x+140 ?

The solutions to the equation x^2+2x=10x+140 are x=2(2+sqrt(39)),x=2(2-sqrt(39))

Find the zeros of x^2+2x=10x+140

The zeros of x^2+2x=10x+140 are x=2(2+sqrt(39)),x=2(2-sqrt(39))

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x^{2} + 2 x = 10 x + 140

x = 2 (2 + 39), x = 2 (2 - 39)

Decimal Notation

x = 16.48999 \dots, x = - 8.48999 \dots

Solution steps

x^{2} + 2 x = 10 x + 140

Move

140

to the left side

x^{2} + 2 x - 140 = 10 x

Move

10 x

to the left side

x^{2} - 8 x - 140 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 8 ) \pm ( - 8 ) ^{2} - 4 \cdot 1 \cdot ( - 140 )}{2 \cdot 1}

(- 8)^{2} - 4 \cdot 1 \cdot (- 140) = 439

x_{1, 2} = \frac{- ( - 8 ) \pm 4 39}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 8 ) + 4 39}{2 \cdot 1}, x_{2} = \frac{- ( - 8 ) - 4 39}{2 \cdot 1}

x = \frac{- ( - 8 ) + 4 39}{2 \cdot 1} : 2 (2 + 39)

x = \frac{- ( - 8 ) - 4 39}{2 \cdot 1} : 2 (2 - 39)

The solutions to the quadratic equation are:

x = 2 (2 + 39), x = 2 (2 - 39)

- 5 n - 4 (- 7 - 4 n) = 36 + 7 n

(x - 4)^{2} = 5

17 x y^{4} + 51 x^{3} y^{3}

12 x^{2} - 12 x = 0

2 (3 t - 1) - 5 t > - 6 (1 - t) + 7

What are the solutions to the equation x^2+2x=10x+140 ?
The solutions to the equation x^2+2x=10x+140 are x=2(2+sqrt(39)),x=2(2-sqrt(39))
Find the zeros of x^2+2x=10x+140
The zeros of x^2+2x=10x+140 are x=2(2+sqrt(39)),x=2(2-sqrt(39))