What are the solutions to the equation x(5x+6)=7 ?

The solutions to the equation x(5x+6)=7 are x=(-3+2sqrt(11))/5 ,x=-(3+2sqrt(11))/5

Find the zeros of x(5x+6)=7

The zeros of x(5x+6)=7 are x=(-3+2sqrt(11))/5 ,x=-(3+2sqrt(11))/5

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x (5 x + 6) = 7

x = \frac{- 3 + 2 11}{5}, x = - \frac{3 + 2 11}{5}

Decimal Notation

x = 0.72664 \dots, x = - 1.92664 \dots

Solution steps

x (5 x + 6) = 7

Expand

x (5 x + 6) : 5 x^{2} + 6 x

5 x^{2} + 6 x = 7

Move

7

to the left side

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

Solve with the quadratic formula

x_{1, 2} = \frac{- 6 \pm 6 ^{2} - 4 \cdot 5 ( - 7 )}{2 \cdot 5}

6^{2} - 4 \cdot 5 (- 7) = 411

x_{1, 2} = \frac{- 6 \pm 4 11}{2 \cdot 5}

Separate the solutions

x_{1} = \frac{- 6 + 4 11}{2 \cdot 5}, x_{2} = \frac{- 6 - 4 11}{2 \cdot 5}

x = \frac{- 6 + 4 11}{2 \cdot 5} : \frac{- 3 + 2 11}{5}

x = \frac{- 6 - 4 11}{2 \cdot 5} : - \frac{3 + 2 11}{5}

The solutions to the quadratic equation are:

x = \frac{- 3 + 2 11}{5}, x = - \frac{3 + 2 11}{5}

x + 11 = 3 x + 1

10 - x = \frac{7 x}{6} + \frac{1}{4}

x^{2} + 4 x + 9 < 0

x^{3} - 8

\frac{15 b}{45 b ^{5}}

What are the solutions to the equation x(5x+6)=7 ?
The solutions to the equation x(5x+6)=7 are x=(-3+2sqrt(11))/5 ,x=-(3+2sqrt(11))/5
Find the zeros of x(5x+6)=7
The zeros of x(5x+6)=7 are x=(-3+2sqrt(11))/5 ,x=-(3+2sqrt(11))/5