What are the solutions to the equation 5x^2=34x+7 ?

The solutions to the equation 5x^2=34x+7 are x=7,x=-1/5

Find the zeros of 5x^2=34x+7

The zeros of 5x^2=34x+7 are x=7,x=-1/5

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5 x^{2} = 34 x + 7

x = 7, x = - \frac{1}{5}

Decimal Notation

x = 7, x = - 0.2

Solution steps

5 x^{2} = 34 x + 7

Move

7

to the left side

5 x^{2} - 7 = 34 x

Move

34 x

to the left side

5 x^{2} - 7 - 34 x = 0

Write in the standard form

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

5 x^{2} - 34 x - 7 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 34 ) \pm ( - 34 ) ^{2} - 4 \cdot 5 ( - 7 )}{2 \cdot 5}

(- 34)^{2} - 4 \cdot 5 (- 7) = 36

x_{1, 2} = \frac{- ( - 34 ) \pm 36}{2 \cdot 5}

Separate the solutions

x_{1} = \frac{- ( - 34 ) + 36}{2 \cdot 5}, x_{2} = \frac{- ( - 34 ) - 36}{2 \cdot 5}

x = \frac{- ( - 34 ) + 36}{2 \cdot 5} : 7

x = \frac{- ( - 34 ) - 36}{2 \cdot 5} : - \frac{1}{5}

The solutions to the quadratic equation are:

x = 7, x = - \frac{1}{5}

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

- \frac{1}{3} (- 4)^{2} + 5

\frac{3}{2} (x - 4) < 2 - \frac{3}{5} (1 + 4 x)

8.99 \cdot 1 0^{9} \cdot \frac{5.4 \cdot 1 0 ^{- 6}}{3.01}

\frac{x}{x + 5} - \frac{4}{x - 15} < 0

What are the solutions to the equation 5x^2=34x+7 ?
The solutions to the equation 5x^2=34x+7 are x=7,x=-1/5
Find the zeros of 5x^2=34x+7
The zeros of 5x^2=34x+7 are x=7,x=-1/5