What are the solutions to the equation 4x^2+7x=5 ?

The solutions to the equation 4x^2+7x=5 are x=(-7+sqrt(129))/8 ,x=(-7-sqrt(129))/8

Find the zeros of 4x^2+7x=5

The zeros of 4x^2+7x=5 are x=(-7+sqrt(129))/8 ,x=(-7-sqrt(129))/8

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4 x^{2} + 7 x = 5

x = \frac{- 7 + 129}{8}, x = \frac{- 7 - 129}{8}

Decimal Notation

x = 0.54472 \dots, x = - 2.29472 \dots

Solution steps

4 x^{2} + 7 x = 5

Move

5

to the left side

4 x^{2} + 7 x - 5 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 7 \pm 7 ^{2} - 4 \cdot 4 ( - 5 )}{2 \cdot 4}

7^{2} - 4 \cdot 4 (- 5) = 129

x_{1, 2} = \frac{- 7 \pm 129}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- 7 + 129}{2 \cdot 4}, x_{2} = \frac{- 7 - 129}{2 \cdot 4}

x = \frac{- 7 + 129}{2 \cdot 4} : \frac{- 7 + 129}{8}

x = \frac{- 7 - 129}{2 \cdot 4} : \frac{- 7 - 129}{8}

The solutions to the quadratic equation are:

x = \frac{- 7 + 129}{8}, x = \frac{- 7 - 129}{8}

\frac{22 + 18}{2}

- 3 ∣ x + 5 ∣ + 1 = 7 ∣ x + 5 ∣ + 8

4 x^{2} - 12 x - 7 = 0

3 (x - 2) = 9

- \frac{3}{4} ∣ x + 5 ∣ + \frac{5}{2} = 1

What are the solutions to the equation 4x^2+7x=5 ?
The solutions to the equation 4x^2+7x=5 are x=(-7+sqrt(129))/8 ,x=(-7-sqrt(129))/8
Find the zeros of 4x^2+7x=5
The zeros of 4x^2+7x=5 are x=(-7+sqrt(129))/8 ,x=(-7-sqrt(129))/8