What are the solutions to the equation 2x^2+14x=3 ?

The solutions to the equation 2x^2+14x=3 are x=(-7+sqrt(55))/2 ,x=-(7+sqrt(55))/2

Find the zeros of 2x^2+14x=3

The zeros of 2x^2+14x=3 are x=(-7+sqrt(55))/2 ,x=-(7+sqrt(55))/2

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2 x^{2} + 14 x = 3

x = \frac{- 7 + 55}{2}, x = - \frac{7 + 55}{2}

Decimal Notation

x = 0.20809 \dots, x = - 7.20809 \dots

Solution steps

2 x^{2} + 14 x = 3

Move

3

to the left side

2 x^{2} + 14 x - 3 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 14 \pm 1 4 ^{2} - 4 \cdot 2 ( - 3 )}{2 \cdot 2}

1 4^{2} - 4 \cdot 2 (- 3) = 255

x_{1, 2} = \frac{- 14 \pm 2 55}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 14 + 2 55}{2 \cdot 2}, x_{2} = \frac{- 14 - 2 55}{2 \cdot 2}

x = \frac{- 14 + 2 55}{2 \cdot 2} : \frac{- 7 + 55}{2}

x = \frac{- 14 - 2 55}{2 \cdot 2} : - \frac{7 + 55}{2}

The solutions to the quadratic equation are:

x = \frac{- 7 + 55}{2}, x = - \frac{7 + 55}{2}

4 - \frac{x}{10} = 8

u^{2} + 13 u + 22

\frac{8}{3} - \frac{x}{2} \leq \frac{2 x}{3} + \frac{1}{3}

∣ 3 x - 4 ∣ \geq 9

1 (10925590.85) (\frac{1}{1} - \frac{1}{4})

What are the solutions to the equation 2x^2+14x=3 ?
The solutions to the equation 2x^2+14x=3 are x=(-7+sqrt(55))/2 ,x=-(7+sqrt(55))/2
Find the zeros of 2x^2+14x=3
The zeros of 2x^2+14x=3 are x=(-7+sqrt(55))/2 ,x=-(7+sqrt(55))/2