What are the solutions to the equation 5x^2-3x-7=0 ?

The solutions to the equation 5x^2-3x-7=0 are x=(3+sqrt(149))/(10),x=(3-sqrt(149))/(10)

Find the zeros of 5x^2-3x-7=0

The zeros of 5x^2-3x-7=0 are x=(3+sqrt(149))/(10),x=(3-sqrt(149))/(10)

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5 x^{2} - 3 x - 7 = 0

x = \frac{3 + 149}{10}, x = \frac{3 - 149}{10}

Decimal Notation

x = 1.52065 \dots, x = - 0.92065 \dots

Solution steps

5 x^{2} - 3 x - 7 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 \cdot 5 ( - 7 )}{2 \cdot 5}

(- 3)^{2} - 4 \cdot 5 (- 7) = 149

x_{1, 2} = \frac{- ( - 3 ) \pm 149}{2 \cdot 5}

Separate the solutions

x_{1} = \frac{- ( - 3 ) + 149}{2 \cdot 5}, x_{2} = \frac{- ( - 3 ) - 149}{2 \cdot 5}

x = \frac{- ( - 3 ) + 149}{2 \cdot 5} : \frac{3 + 149}{10}

x = \frac{- ( - 3 ) - 149}{2 \cdot 5} : \frac{3 - 149}{10}

The solutions to the quadratic equation are:

x = \frac{3 + 149}{10}, x = \frac{3 - 149}{10}

\frac{13}{30} - \frac{1}{30}

3 (5 x + 4) = 4 (2 - 3 x)

(x^{- 6})^{\frac{5}{3}}

0.5 x^{2} + 1 \leq 5 ∣ x ∣

\frac{- 3}{- 6}

What are the solutions to the equation 5x^2-3x-7=0 ?
The solutions to the equation 5x^2-3x-7=0 are x=(3+sqrt(149))/(10),x=(3-sqrt(149))/(10)
Find the zeros of 5x^2-3x-7=0
The zeros of 5x^2-3x-7=0 are x=(3+sqrt(149))/(10),x=(3-sqrt(149))/(10)