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

The solutions to the equation 4x^2-3=-4+7x are x=(7+sqrt(33))/8 ,x=(7-sqrt(33))/8

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

The zeros of 4x^2-3=-4+7x are x=(7+sqrt(33))/8 ,x=(7-sqrt(33))/8

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

x = \frac{7 + 33}{8}, x = \frac{7 - 33}{8}

Solution steps

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

Move

7 x

to the left side

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

Move

4

to the left side

4 x^{2} - 7 x + 1 = 0

Solve with the quadratic formula

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

(- 7)^{2} - 4 \cdot 4 \cdot 1 = 33

x_{1, 2} = \frac{- ( - 7 ) \pm 33}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- ( - 7 ) + 33}{2 \cdot 4}, x_{2} = \frac{- ( - 7 ) - 33}{2 \cdot 4}

x = \frac{- ( - 7 ) + 33}{2 \cdot 4} : \frac{7 + 33}{8}

x = \frac{- ( - 7 ) - 33}{2 \cdot 4} : \frac{7 - 33}{8}

The solutions to the quadratic equation are:

x = \frac{7 + 33}{8}, x = \frac{7 - 33}{8}

2 x^{3} - 2 x^{2} - 112 x

2 n^{2} + 2 n - 4 = 0

\frac{\frac{1}{6 + x + h} - \frac{1}{6 + x}}{h}

g - 16 = 4

2 x^{2} = 9 x + 5

What are the solutions to the equation 4x^2-3=-4+7x ?
The solutions to the equation 4x^2-3=-4+7x are x=(7+sqrt(33))/8 ,x=(7-sqrt(33))/8
Find the zeros of 4x^2-3=-4+7x
The zeros of 4x^2-3=-4+7x are x=(7+sqrt(33))/8 ,x=(7-sqrt(33))/8