What are the solutions to the equation 2x(6-5x)=-7 ?

The solutions to the equation 2x(6-5x)=-7 are x=-(-6+sqrt(106))/(10),x=(6+sqrt(106))/(10)

Find the zeros of 2x(6-5x)=-7

The zeros of 2x(6-5x)=-7 are x=-(-6+sqrt(106))/(10),x=(6+sqrt(106))/(10)

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2 x (6 - 5 x) = - 7

x = - \frac{- 6 + 106}{10}, x = \frac{6 + 106}{10}

Decimal Notation

x = - 0.42956 \dots, x = 1.62956 \dots

Solution steps

2 x (6 - 5 x) = - 7

Expand

2 x (6 - 5 x) : 12 x - 10 x^{2}

12 x - 10 x^{2} = - 7

Move

7

to the left side

12 x - 10 x^{2} + 7 = 0

Write in the standard form

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

- 10 x^{2} + 12 x + 7 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 12 \pm 1 2 ^{2} - 4 ( - 10 ) \cdot 7}{2 ( - 10 )}

1 2^{2} - 4 (- 10) \cdot 7 = 2106

x_{1, 2} = \frac{- 12 \pm 2 106}{2 ( - 10 )}

Separate the solutions

x_{1} = \frac{- 12 + 2 106}{2 ( - 10 )}, x_{2} = \frac{- 12 - 2 106}{2 ( - 10 )}

x = \frac{- 12 + 2 106}{2 ( - 10 )} : - \frac{- 6 + 106}{10}

x = \frac{- 12 - 2 106}{2 ( - 10 )} : \frac{6 + 106}{10}

The solutions to the quadratic equation are:

x = - \frac{- 6 + 106}{10}, x = \frac{6 + 106}{10}

- 3 r^{2} - r - 10 = - 4 r^{2}

x^{2} + 10 x + 25 = 0

2 x^{2} + 4 = - 7 x

8 q^{2} - 26 q + 15 = 0

r^{2} + 1

What are the solutions to the equation 2x(6-5x)=-7 ?
The solutions to the equation 2x(6-5x)=-7 are x=-(-6+sqrt(106))/(10),x=(6+sqrt(106))/(10)
Find the zeros of 2x(6-5x)=-7
The zeros of 2x(6-5x)=-7 are x=-(-6+sqrt(106))/(10),x=(6+sqrt(106))/(10)