What are the solutions to the equation 2x(3x-5)+7x(1-x)=-10 ?

The solutions to the equation 2x(3x-5)+7x(1-x)=-10 are x=-5,x=2

Find the zeros of 2x(3x-5)+7x(1-x)=-10

The zeros of 2x(3x-5)+7x(1-x)=-10 are x=-5,x=2

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2 x (3 x - 5) + 7 x (1 - x) = - 10

x = - 5, x = 2

Solution steps

2 x (3 x - 5) + 7 x (1 - x) = - 10

Expand

2 x (3 x - 5) + 7 x (1 - x) : - x^{2} - 3 x

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

Move

10

to the left side

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

Solve with the quadratic formula

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

(- 3)^{2} - 4 (- 1) \cdot 10 = 7

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

Separate the solutions

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

x = \frac{- ( - 3 ) + 7}{2 ( - 1 )} : - 5

x = \frac{- ( - 3 ) - 7}{2 ( - 1 )} : 2

The solutions to the quadratic equation are:

x = - 5, x = 2

x^{2} - \frac{3}{2} x + 3 = 0

x^{- \frac{3}{2}}

8 t = - 2

5 = v - 6

What are the solutions to the equation 2x(3x-5)+7x(1-x)=-10 ?
The solutions to the equation 2x(3x-5)+7x(1-x)=-10 are x=-5,x=2
Find the zeros of 2x(3x-5)+7x(1-x)=-10
The zeros of 2x(3x-5)+7x(1-x)=-10 are x=-5,x=2