What are the solutions to the equation 4x^2+42x=2x-8 ?

The solutions to the equation 4x^2+42x=2x-8 are x=-5+sqrt(23),x=-5-sqrt(23)

Find the zeros of 4x^2+42x=2x-8

The zeros of 4x^2+42x=2x-8 are x=-5+sqrt(23),x=-5-sqrt(23)

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4 x^{2} + 42 x = 2 x - 8

x = - 5 + 23, x = - 5 - 23

Decimal Notation

x = - 0.20416 \dots, x = - 9.79583 \dots

Solution steps

4 x^{2} + 42 x = 2 x - 8

Move

8

to the left side

4 x^{2} + 42 x + 8 = 2 x

Move

2 x

to the left side

4 x^{2} + 40 x + 8 = 0

Solve with the quadratic formula

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

4 0^{2} - 4 \cdot 4 \cdot 8 = 823

x_{1, 2} = \frac{- 40 \pm 8 23}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- 40 + 8 23}{2 \cdot 4}, x_{2} = \frac{- 40 - 8 23}{2 \cdot 4}

x = \frac{- 40 + 8 23}{2 \cdot 4} : - 5 + 23

x = \frac{- 40 - 8 23}{2 \cdot 4} : - 5 - 23

The solutions to the quadratic equation are:

x = - 5 + 23, x = - 5 - 23

\frac{1}{2} r + 3 < - 6

\frac{3 ^{8}}{3 ^{2}}

2 - \frac{6}{5} x \geq - 4

5^{2} = 1^{2} + x^{2}

g^{3} - 343

What are the solutions to the equation 4x^2+42x=2x-8 ?
The solutions to the equation 4x^2+42x=2x-8 are x=-5+sqrt(23),x=-5-sqrt(23)
Find the zeros of 4x^2+42x=2x-8
The zeros of 4x^2+42x=2x-8 are x=-5+sqrt(23),x=-5-sqrt(23)