What are the solutions to the equation 299x^2+100x=500-101x^2 ?

The solutions to the equation 299x^2+100x=500-101x^2 are x=1,x=-5/4

Find the zeros of 299x^2+100x=500-101x^2

The zeros of 299x^2+100x=500-101x^2 are x=1,x=-5/4

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299 x^{2} + 100 x = 500 - 101 x^{2}

x = 1, x = - \frac{5}{4}

Decimal Notation

x = 1, x = - 1.25

Solution steps

299 x^{2} + 100 x = 500 - 101 x^{2}

Move

101 x^{2}

to the left side

400 x^{2} + 100 x = 500

Move

500

to the left side

400 x^{2} + 100 x - 500 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 100 \pm 10 0 ^{2} - 4 \cdot 400 ( - 500 )}{2 \cdot 400}

10 0^{2} - 4 \cdot 400 (- 500) = 900

x_{1, 2} = \frac{- 100 \pm 900}{2 \cdot 400}

Separate the solutions

x_{1} = \frac{- 100 + 900}{2 \cdot 400}, x_{2} = \frac{- 100 - 900}{2 \cdot 400}

x = \frac{- 100 + 900}{2 \cdot 400} : 1

x = \frac{- 100 - 900}{2 \cdot 400} : - \frac{5}{4}

The solutions to the quadratic equation are:

x = 1, x = - \frac{5}{4}

4 x + 3 \geq 7 x - 1

s^{3} + 2 s^{2} + s

\frac{- 1.38 \cdot 303.15}{- 1.6 \cdot 1 0 ^{- 4}}

5 y + 6 y - 81 = 7 y + 102 + 65 y

8 (0) - 2 (7) \geq - 12

What are the solutions to the equation 299x^2+100x=500-101x^2 ?
The solutions to the equation 299x^2+100x=500-101x^2 are x=1,x=-5/4
Find the zeros of 299x^2+100x=500-101x^2
The zeros of 299x^2+100x=500-101x^2 are x=1,x=-5/4