What are the solutions to the equation 3x^2-5(2x+7)=0 ?

The solutions to the equation 3x^2-5(2x+7)=0 are x=(5+sqrt(130))/3 ,x=(5-sqrt(130))/3

Find the zeros of 3x^2-5(2x+7)=0

The zeros of 3x^2-5(2x+7)=0 are x=(5+sqrt(130))/3 ,x=(5-sqrt(130))/3

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3 x^{2} - 5 (2 x + 7) = 0

x = \frac{5 + 130}{3}, x = \frac{5 - 130}{3}

Solution steps

3 x^{2} - 5 (2 x + 7) = 0

Expand

3 x^{2} - 5 (2 x + 7) : 3 x^{2} - 10 x - 35

3 x^{2} - 10 x - 35 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 10 ) \pm ( - 10 ) ^{2} - 4 \cdot 3 ( - 35 )}{2 \cdot 3}

(- 10)^{2} - 4 \cdot 3 (- 35) = 2130

x_{1, 2} = \frac{- ( - 10 ) \pm 2 130}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- ( - 10 ) + 2 130}{2 \cdot 3}, x_{2} = \frac{- ( - 10 ) - 2 130}{2 \cdot 3}

x = \frac{- ( - 10 ) + 2 130}{2 \cdot 3} : \frac{5 + 130}{3}

x = \frac{- ( - 10 ) - 2 130}{2 \cdot 3} : \frac{5 - 130}{3}

The solutions to the quadratic equation are:

x = \frac{5 + 130}{3}, x = \frac{5 - 130}{3}

3 b^{2} - b + 5 = 6

x, (2 x - 9 y)^{2} + 7 y = 27 x

3 y^{2} + 2 = 5 y

\frac{( 2 x + 3 ) ^{2}}{2} = - 1

9 x^{2} - 8 = 0

What are the solutions to the equation 3x^2-5(2x+7)=0 ?
The solutions to the equation 3x^2-5(2x+7)=0 are x=(5+sqrt(130))/3 ,x=(5-sqrt(130))/3
Find the zeros of 3x^2-5(2x+7)=0
The zeros of 3x^2-5(2x+7)=0 are x=(5+sqrt(130))/3 ,x=(5-sqrt(130))/3