What are the solutions to the equation 300=t^2+2t ?

The solutions to the equation 300=t^2+2t are t=-1+sqrt(301),t=-1-sqrt(301)

Find the zeros of 300=t^2+2t

The zeros of 300=t^2+2t are t=-1+sqrt(301),t=-1-sqrt(301)

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300 = t^{2} + 2 t

t = - 1 + 301, t = - 1 - 301

Decimal Notation

t = 16.34935 \dots, t = - 18.34935 \dots

Solution steps

300 = t^{2} + 2 t

Switch sides

t^{2} + 2 t = 300

Move

300

to the left side

t^{2} + 2 t - 300 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 \cdot 1 \cdot ( - 300 )}{2 \cdot 1}

2^{2} - 4 \cdot 1 \cdot (- 300) = 2301

t_{1, 2} = \frac{- 2 \pm 2 301}{2 \cdot 1}

Separate the solutions

t_{1} = \frac{- 2 + 2 301}{2 \cdot 1}, t_{2} = \frac{- 2 - 2 301}{2 \cdot 1}

t = \frac{- 2 + 2 301}{2 \cdot 1} : - 1 + 301

t = \frac{- 2 - 2 301}{2 \cdot 1} : - 1 - 301

The solutions to the quadratic equation are:

t = - 1 + 301, t = - 1 - 301

- x^{2} + 21 x - 74 = 0

x, 2 y + 6 x = 3 x^{2} - y^{4}

- 2 x^{2} - 3 x + 20 = 0

x^{2} - 2 (x - 6) = 4 x + 4

6 (5 x + x)^{2} - 42 = 0

What are the solutions to the equation 300=t^2+2t ?
The solutions to the equation 300=t^2+2t are t=-1+sqrt(301),t=-1-sqrt(301)
Find the zeros of 300=t^2+2t
The zeros of 300=t^2+2t are t=-1+sqrt(301),t=-1-sqrt(301)