What are the solutions to the equation 5x^2+39x-125=0 ?

The solutions to the equation 5x^2+39x-125=0 are x=(-39+sqrt(4021))/(10),x=(-39-sqrt(4021))/(10)

Find the zeros of 5x^2+39x-125=0

The zeros of 5x^2+39x-125=0 are x=(-39+sqrt(4021))/(10),x=(-39-sqrt(4021))/(10)

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5 x^{2} + 39 x - 125 = 0

x = \frac{- 39 + 4021}{10}, x = \frac{- 39 - 4021}{10}

Decimal Notation

x = 2.44113 \dots, x = - 10.24113 \dots

Solution steps

5 x^{2} + 39 x - 125 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 39 \pm 3 9 ^{2} - 4 \cdot 5 ( - 125 )}{2 \cdot 5}

3 9^{2} - 4 \cdot 5 (- 125) = 4021

x_{1, 2} = \frac{- 39 \pm 4021}{2 \cdot 5}

Separate the solutions

x_{1} = \frac{- 39 + 4021}{2 \cdot 5}, x_{2} = \frac{- 39 - 4021}{2 \cdot 5}

x = \frac{- 39 + 4021}{2 \cdot 5} : \frac{- 39 + 4021}{10}

x = \frac{- 39 - 4021}{2 \cdot 5} : \frac{- 39 - 4021}{10}

The solutions to the quadratic equation are:

x = \frac{- 39 + 4021}{10}, x = \frac{- 39 - 4021}{10}

x, y = - x^{2} + 10 x - 18

(X - 15)^{2} + (X + 4 + 7)^{2} = 9

296 = 4 x^{2} + 140 x

C = 200 - 7 x + 0.345 x^{2}

x^{2} = 23 x

What are the solutions to the equation 5x^2+39x-125=0 ?
The solutions to the equation 5x^2+39x-125=0 are x=(-39+sqrt(4021))/(10),x=(-39-sqrt(4021))/(10)
Find the zeros of 5x^2+39x-125=0
The zeros of 5x^2+39x-125=0 are x=(-39+sqrt(4021))/(10),x=(-39-sqrt(4021))/(10)