What are the solutions to the equation 4x^2+24x-68=0 ?

The solutions to the equation 4x^2+24x-68=0 are x=-3+sqrt(26),x=-3-sqrt(26)

Find the zeros of 4x^2+24x-68=0

The zeros of 4x^2+24x-68=0 are x=-3+sqrt(26),x=-3-sqrt(26)

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4 x^{2} + 24 x - 68 = 0

x = - 3 + 26, x = - 3 - 26

Decimal Notation

x = 2.09901 \dots, x = - 8.09901 \dots

Solution steps

4 x^{2} + 24 x - 68 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 24 \pm 2 4 ^{2} - 4 \cdot 4 ( - 68 )}{2 \cdot 4}

2 4^{2} - 4 \cdot 4 (- 68) = 826

x_{1, 2} = \frac{- 24 \pm 8 26}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- 24 + 8 26}{2 \cdot 4}, x_{2} = \frac{- 24 - 8 26}{2 \cdot 4}

x = \frac{- 24 + 8 26}{2 \cdot 4} : - 3 + 26

x = \frac{- 24 - 8 26}{2 \cdot 4} : - 3 - 26

The solutions to the quadratic equation are:

x = - 3 + 26, x = - 3 - 26

- 1 > (\frac{1}{x - 1})

\frac{x ^{2}}{x ^{3}}

\frac{n + 4}{10 n ^{2} + 40 n}

x^{3} - 343

- 23 = y - 18

What are the solutions to the equation 4x^2+24x-68=0 ?
The solutions to the equation 4x^2+24x-68=0 are x=-3+sqrt(26),x=-3-sqrt(26)
Find the zeros of 4x^2+24x-68=0
The zeros of 4x^2+24x-68=0 are x=-3+sqrt(26),x=-3-sqrt(26)