What are the solutions to the equation 4x^2+7=29x ?

The solutions to the equation 4x^2+7=29x are x=7,x= 1/4

Find the zeros of 4x^2+7=29x

The zeros of 4x^2+7=29x are x=7,x= 1/4

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4 x^{2} + 7 = 29 x

x = 7, x = \frac{1}{4}

Solution steps

4 x^{2} + 7 = 29 x

Move

29 x

to the left side

4 x^{2} + 7 - 29 x = 0

Write in the standard form

a x^{2} + b x + c = 0

4 x^{2} - 29 x + 7 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 29 ) \pm ( - 29 ) ^{2} - 4 \cdot 4 \cdot 7}{2 \cdot 4}

(- 29)^{2} - 4 \cdot 4 \cdot 7 = 27

x_{1, 2} = \frac{- ( - 29 ) \pm 27}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- ( - 29 ) + 27}{2 \cdot 4}, x_{2} = \frac{- ( - 29 ) - 27}{2 \cdot 4}

x = \frac{- ( - 29 ) + 27}{2 \cdot 4} : 7

x = \frac{- ( - 29 ) - 27}{2 \cdot 4} : \frac{1}{4}

The solutions to the quadratic equation are:

x = 7, x = \frac{1}{4}

i^{25}

(2 x + 3) (1 - 3 x) (x - 2) < 0

3 x^{2} - 15 x = 2 x + 6

(x + 3) + 0 + (5 x) = 15

x^{4} - 36

What are the solutions to the equation 4x^2+7=29x ?
The solutions to the equation 4x^2+7=29x are x=7,x= 1/4
Find the zeros of 4x^2+7=29x
The zeros of 4x^2+7=29x are x=7,x= 1/4