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

The solutions to the equation 5x^2+5x-3=2x^2 are x=(-5+sqrt(61))/6 ,x=(-5-sqrt(61))/6

Find the zeros of 5x^2+5x-3=2x^2

The zeros of 5x^2+5x-3=2x^2 are x=(-5+sqrt(61))/6 ,x=(-5-sqrt(61))/6

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5 x^{2} + 5 x - 3 = 2 x^{2}

x = \frac{- 5 + 61}{6}, x = \frac{- 5 - 61}{6}

Solution steps

5 x^{2} + 5 x - 3 = 2 x^{2}

Move

2 x^{2}

to the left side

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

Solve with the quadratic formula

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

5^{2} - 4 \cdot 3 (- 3) = 61

x_{1, 2} = \frac{- 5 \pm 61}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- 5 + 61}{2 \cdot 3}, x_{2} = \frac{- 5 - 61}{2 \cdot 3}

x = \frac{- 5 + 61}{2 \cdot 3} : \frac{- 5 + 61}{6}

x = \frac{- 5 - 61}{2 \cdot 3} : \frac{- 5 - 61}{6}

The solutions to the quadratic equation are:

x = \frac{- 5 + 61}{6}, x = \frac{- 5 - 61}{6}

\frac{( \frac{1}{4} x ^{2} + a x + ( 2 a - 8 ) - 8 )}{2 ( x + 2 )}

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

16.5 \cdot \frac{1}{10}

8 x < 24 or x - 4 > 1

\frac{5 k ^{2} - 50 k + 45}{k ^{2} - 18 k + 81}

What are the solutions to the equation 5x^2+5x-3=2x^2 ?
The solutions to the equation 5x^2+5x-3=2x^2 are x=(-5+sqrt(61))/6 ,x=(-5-sqrt(61))/6
Find the zeros of 5x^2+5x-3=2x^2
The zeros of 5x^2+5x-3=2x^2 are x=(-5+sqrt(61))/6 ,x=(-5-sqrt(61))/6