What are the solutions to the equation 5x^2-7x=-4+3x^2 ?

The solutions to the equation 5x^2-7x=-4+3x^2 are x=(7+sqrt(17))/4 ,x=(7-sqrt(17))/4

Find the zeros of 5x^2-7x=-4+3x^2

The zeros of 5x^2-7x=-4+3x^2 are x=(7+sqrt(17))/4 ,x=(7-sqrt(17))/4

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5 x^{2} - 7 x = - 4 + 3 x^{2}

x = \frac{7 + 17}{4}, x = \frac{7 - 17}{4}

Decimal Notation

x = 2.78077 \dots, x = 0.71922 \dots

Solution steps

5 x^{2} - 7 x = - 4 + 3 x^{2}

Move

3 x^{2}

to the left side

2 x^{2} - 7 x = - 4

Move

4

to the left side

2 x^{2} - 7 x + 4 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 7 ) \pm ( - 7 ) ^{2} - 4 \cdot 2 \cdot 4}{2 \cdot 2}

(- 7)^{2} - 4 \cdot 2 \cdot 4 = 17

x_{1, 2} = \frac{- ( - 7 ) \pm 17}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- ( - 7 ) + 17}{2 \cdot 2}, x_{2} = \frac{- ( - 7 ) - 17}{2 \cdot 2}

x = \frac{- ( - 7 ) + 17}{2 \cdot 2} : \frac{7 + 17}{4}

x = \frac{- ( - 7 ) - 17}{2 \cdot 2} : \frac{7 - 17}{4}

The solutions to the quadratic equation are:

x = \frac{7 + 17}{4}, x = \frac{7 - 17}{4}

- 11 a^{2} + 32 a + 256 = 0

- 7 x^{2} - 21 x - 5

3 x + 5 = 4 x^{2}

t, 2 a t^{2} - 6 a t - a + 2 t b - 3 b + 2 c = t^{2} + 3

2 \cdot ((0.145 - 2 x)^{2}) = (0.0262 + x) \cdot (0.074 + x)

What are the solutions to the equation 5x^2-7x=-4+3x^2 ?
The solutions to the equation 5x^2-7x=-4+3x^2 are x=(7+sqrt(17))/4 ,x=(7-sqrt(17))/4
Find the zeros of 5x^2-7x=-4+3x^2
The zeros of 5x^2-7x=-4+3x^2 are x=(7+sqrt(17))/4 ,x=(7-sqrt(17))/4