What are the solutions to the equation 5x^2=16+7x ?

The solutions to the equation 5x^2=16+7x are x=(7+3sqrt(41))/(10),x=(7-3sqrt(41))/(10)

Find the zeros of 5x^2=16+7x

The zeros of 5x^2=16+7x are x=(7+3sqrt(41))/(10),x=(7-3sqrt(41))/(10)

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5 x^{2} = 16 + 7 x

x = \frac{7 + 3 41}{10}, x = \frac{7 - 3 41}{10}

Solution steps

5 x^{2} = 16 + 7 x

Move

7 x

to the left side

5 x^{2} - 7 x = 16

Move

16

to the left side

5 x^{2} - 7 x - 16 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 7 ) \pm ( - 7 ) ^{2} - 4 \cdot 5 ( - 16 )}{2 \cdot 5}

(- 7)^{2} - 4 \cdot 5 (- 16) = 341

x_{1, 2} = \frac{- ( - 7 ) \pm 3 41}{2 \cdot 5}

Separate the solutions

x_{1} = \frac{- ( - 7 ) + 3 41}{2 \cdot 5}, x_{2} = \frac{- ( - 7 ) - 3 41}{2 \cdot 5}

x = \frac{- ( - 7 ) + 3 41}{2 \cdot 5} : \frac{7 + 3 41}{10}

x = \frac{- ( - 7 ) - 3 41}{2 \cdot 5} : \frac{7 - 3 41}{10}

The solutions to the quadratic equation are:

x = \frac{7 + 3 41}{10}, x = \frac{7 - 3 41}{10}

6^{2} + 5^{2}

28 x^{3} y^{3}

\frac{2 x + 1}{7} > \frac{x - 3}{4}

a^{2} + 2 a + 1

49 - (x + 3)^{2}

What are the solutions to the equation 5x^2=16+7x ?
The solutions to the equation 5x^2=16+7x are x=(7+3sqrt(41))/(10),x=(7-3sqrt(41))/(10)
Find the zeros of 5x^2=16+7x
The zeros of 5x^2=16+7x are x=(7+3sqrt(41))/(10),x=(7-3sqrt(41))/(10)