What are the solutions to the equation 15x-(11x+5)=-x^2 ?

The solutions to the equation 15x-(11x+5)=-x^2 are x=-5,x=1

Find the zeros of 15x-(11x+5)=-x^2

The zeros of 15x-(11x+5)=-x^2 are x=-5,x=1

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15 x - (11 x + 5) = - x^{2}

x = - 5, x = 1

Solution steps

15 x - (11 x + 5) = - x^{2}

Expand

15 x - (11 x + 5) : 4 x - 5

4 x - 5 = - x^{2}

Switch sides

- x^{2} = 4 x - 5

Move

5

to the left side

- x^{2} + 5 = 4 x

Move

4 x

to the left side

- x^{2} + 5 - 4 x = 0

Write in the standard form

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

- x^{2} - 4 x + 5 = 0

Solve with the quadratic formula

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

(- 4)^{2} - 4 (- 1) \cdot 5 = 6

x_{1, 2} = \frac{- ( - 4 ) \pm 6}{2 ( - 1 )}

Separate the solutions

x_{1} = \frac{- ( - 4 ) + 6}{2 ( - 1 )}, x_{2} = \frac{- ( - 4 ) - 6}{2 ( - 1 )}

x = \frac{- ( - 4 ) + 6}{2 ( - 1 )} : - 5

x = \frac{- ( - 4 ) - 6}{2 ( - 1 )} : 1

The solutions to the quadratic equation are:

x = - 5, x = 1

x^{2} - 8 x = - 18

3 x^{2} + 2 = 14

x^{2} + 4^{2} = 9^{2}

4 (2 x + 4)^{2} + 42 = 46

x, x^{2} + x y - y^{3} = - 1

What are the solutions to the equation 15x-(11x+5)=-x^2 ?
The solutions to the equation 15x-(11x+5)=-x^2 are x=-5,x=1
Find the zeros of 15x-(11x+5)=-x^2
The zeros of 15x-(11x+5)=-x^2 are x=-5,x=1