What are the solutions to the equation h^2+5h=295 ?

The solutions to the equation h^2+5h=295 are h=(-5+sqrt(1205))/2 ,h=(-5-sqrt(1205))/2

Find the zeros of h^2+5h=295

The zeros of h^2+5h=295 are h=(-5+sqrt(1205))/2 ,h=(-5-sqrt(1205))/2

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h^{2} + 5 h = 295

h = \frac{- 5 + 1205}{2}, h = \frac{- 5 - 1205}{2}

Decimal Notation

h = 14.85655 \dots, h = - 19.85655 \dots

Solution steps

h^{2} + 5 h = 295

Move

295

to the left side

h^{2} + 5 h - 295 = 0

Solve with the quadratic formula

h_{1, 2} = \frac{- 5 \pm 5 ^{2} - 4 \cdot 1 \cdot ( - 295 )}{2 \cdot 1}

5^{2} - 4 \cdot 1 \cdot (- 295) = 1205

h_{1, 2} = \frac{- 5 \pm 1205}{2 \cdot 1}

Separate the solutions

h_{1} = \frac{- 5 + 1205}{2 \cdot 1}, h_{2} = \frac{- 5 - 1205}{2 \cdot 1}

h = \frac{- 5 + 1205}{2 \cdot 1} : \frac{- 5 + 1205}{2}

h = \frac{- 5 - 1205}{2 \cdot 1} : \frac{- 5 - 1205}{2}

The solutions to the quadratic equation are:

h = \frac{- 5 + 1205}{2}, h = \frac{- 5 - 1205}{2}

16 z^{2} - 40 z + 31 = 6

- 0.25 x^{2} + 6 x + 20 = 0

y, x^{2} y^{2} + 5 x y = 14

6 x^{2} + 24 x - 126 = 0

4 = 3 x^{2} - 14 x + 4

What are the solutions to the equation h^2+5h=295 ?
The solutions to the equation h^2+5h=295 are h=(-5+sqrt(1205))/2 ,h=(-5-sqrt(1205))/2
Find the zeros of h^2+5h=295
The zeros of h^2+5h=295 are h=(-5+sqrt(1205))/2 ,h=(-5-sqrt(1205))/2