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

The solutions to the equation quadraticformula 3x^2-7x=5 are x=(7+sqrt(109))/6 ,x=(7-sqrt(109))/6

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

The zeros of quadraticformula 3x^2-7x=5 are x=(7+sqrt(109))/6 ,x=(7-sqrt(109))/6

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3 x^{2} - 7 x = 5

x = \frac{7 + 109}{6}, x = \frac{7 - 109}{6}

Decimal Notation

x = 2.90671 \dots, x = - 0.57338 \dots

Solution steps

3 x^{2} - 7 x = 5

Move

5

to the left side

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

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 7 ) \pm ( - 7 ) ^{2} - 4 \cdot 3 ( - 5 )}{2 \cdot 3}

(- 7)^{2} - 4 \cdot 3 (- 5) = 109

x_{1, 2} = \frac{- ( - 7 ) \pm 109}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- ( - 7 ) + 109}{2 \cdot 3}, x_{2} = \frac{- ( - 7 ) - 109}{2 \cdot 3}

x = \frac{- ( - 7 ) + 109}{2 \cdot 3} : \frac{7 + 109}{6}

x = \frac{- ( - 7 ) - 109}{2 \cdot 3} : \frac{7 - 109}{6}

The solutions to the quadratic equation are:

x = \frac{7 + 109}{6}, x = \frac{7 - 109}{6}

d, - 38 d - 57 \geq - 19 d + 76

2 cos (x)

x, \frac{x}{4} = \frac{( x + 9 )}{6.25}

29 = 3

3 \frac{3}{5} /3

What are the solutions to the equation quadraticformula 3x^2-7x=5 ?
The solutions to the equation quadraticformula 3x^2-7x=5 are x=(7+sqrt(109))/6 ,x=(7-sqrt(109))/6
Find the zeros of quadraticformula 3x^2-7x=5
The zeros of quadraticformula 3x^2-7x=5 are x=(7+sqrt(109))/6 ,x=(7-sqrt(109))/6